# Full text of "Plane and spherical trigonometry, and Four-place tables of logarithms"

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■If. 1 ilif 1 TRIGC 1 liWuiRjuu'JunMflwwinBpQOM ffTtfWWKWMWQQWmim yWnTBBtmffllDCCTHHBlUOwHM Q-/45-3 CORNELL UNIVERSITY LIBRARY Given to the COLLEGE OF ENGINEERING by* Mr. J. R. Geib. Date Due ^gbgy^ -m TH38fr Cornell University Library QA 531.G76 Plane and spherical trigonometry, and Fou 3 1924 004 031 708 Cornell University Library The original of this book is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31 924004031 708 MATHEMATICAL TEXTS FOR SCHOOLS AND COLLEGES EDITED BY PERCEY F. SMITH, Ph.D. PROFESSOK OF MATHEMATICS IN THE SHEFFIELD SCIENTIFIC SCHOOL OF YALE UNIVERSITY PLANE AND SPHERICAL TRIGONOMETRY AND FOUR-PLACE TABLES OF LOGARITHMS BY WILLIAM ANTHONY GRANVILLE, Ph.D., LL.D. PRESIDENT OF PENNSYLVANIA COLLEGE GINN AND COMPANY BOSTON • NEW YORK ■ CHICAGO • LONDON ATLANTA • DALLAS • COLUMBUS ■ SAN FRANCISCO Entered at Statioxers' Hall COPYRIGIIT, 1009, BY WILLIAM ANTHONY GRANVILLE ALL EIGHTS RESERVED 815.10 GTNN AND COMPANY ■ PRO- PRIETORS • BOSTON • U.S.A. PREFACE It has been the author's aim to treat the subject according to the latest and most approved methods. The book is designed for the use of colleges, technical schools, normal schools, secondary schools, and for those who take up the subject without the aid of a teacher. Special attention has been paid to the requirements of the College Entrance Board. The book contains more material than is required for some first courses in Trigonometry, but the matter has been so arranged that the teacher can make such omissions as will suit his particular needs. The trigonometric functions are defined as ratios ; first for acute angles in right triangles, and then these definitions are extended to angles in general by means of coordinates. The student is first taught to use the natural functions of acute angles in the solution of simple problems involving right triangles. Attention is called to the methods shown in §§ 23-29 for the reduction of functions of angles outside of the first quadrant. In general, the first example! given under each topic are worked out, making use of the natural functions. A large number of carefully graded exercises are given, and the processes involved are summarized into working rules wherever practicable. Illustrative examples are worked out in detail under each topic. Logarithms are introduced as a separate topic, and attention is called to the fact that they serve to minimize the labor of com- putation. Granville's Four-Place Tables of Logarithms is used. While no radical changes in the usual arrangement of logarithmic tables have been made, several improvements have been effected which greatly facilitate logarithmic computations. Particularly important is the fact that the degree of accuracy which may be expected in a result found by the aid of these tables is clearly indicated. Under each case in the solution of triangles are given two complete sets of examples, — ■ one in which the angles are ex- pressed in degrees and minutes, and another in which the angles are expressed in degrees and the decimal part of a degree. This arrangement, which is characteristic of this book, should be of great vi PREFACE advantage to those secondary schools in which college preparation involving both systems is necessary. To facilitate the drawing of figures and the graphical checking of results a combined ruler and protractor of celluloid is furnished with each copy of the book, and will be found on the inside of the back cover. In Spherical Trigonometry some simplifications have been intro- duced in the application of Napier's rule of circular parts to the solution of right spherical triangles. The treatment of oblique spherical triangles is unique. By making use of the Principle of Duality nearly one half of the work usually required in deriving the standard formulas is done away with, and the usual six cases in the solution of oblique spherical triangles have been reduced to three. An attempt has been made to treat the most important appli- cations of Spherical Trigonometry to Geodesy, Astronomy, and Navi- gation with more clearness and simplicity than has been the rule in elementary treatises. The author's acknowledgments are due to Professor John C. Tracy for many valuable suggestions in the treatment of Spherical Trigo- nometry, to Messrs. L. E. Armstrong and C. C. Perkins for verifying the answers to the problems, and to Mr. S. J. Berard for drawing the figures. W. A. GRANVILLE CONTENTS PLANE TRIGONOMETRY CHAPTER I TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES. SOLUTION OF EIGHT TRIANGLES Section Page 1; Trigonometric functions of an acute angle defined 1 2. Functions of 45°, 30°, 60° ... . ... . 4 3. Solution of right triangles ........ 7 4. General directions for solving right triangles .... .... 7 5. Solution of isosceles triangles . . . . . ... 13 6. Solution of regular polygons . . ... 14 7. Interpolation 16 8. Terms occurring in trigonometric problems 19 CHAPTER II TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 9. Generation of angles ... . . 24 10. Positive and negative angles 24 11. Angles of any magnitude ... . .... 25 12. The four quadrants 25 13. Rectangular coordinates of a point in a plane . . .... 26 14. Distance of a point fro m the origin .... ... . . 27 15. Trigonometric functions of any angle defined .... ... 28 16. Algebraic signs of the trigonometric functions 29 17. Given the value of a function, to construct the angle 29 18. Eive of the trigonometric functions expressed in terms of the sixth . 34 19. Line definitions of the t rigonometric functions .36 20. Changes in the valuesoFTthe functions as the angle varies 38 21. Angular measure . 43 22. Circular measure . . . . . 43 23. Reduction of trigonometric functions to functions of acute angles . 47 24. Functions of complementary angles . . 47 25. Reduction of functions of angles in the second quadrant 47 26. Reduction of functions of angles in the third quadrant .... 51 27. Reduction of functions of angles in the fourth quadrant 53 28. Reduction of functions of negative angles 55 29. General rule for reducing the functions of any angle ...... 57 viii PLANE TKIGONOMETKY CHAPTER III RELATIONS BETWEEN THE TRIGONOMETKIC FUNCTIONS Section Page 30. Fundamental relations between the functions 59 31. Any function expressed in terms of each of the other five functions . 60 CHAPTER IV TRIGONOMETRIC ANALYSIS 32. Functions of the sum and of the difference of two angles 63 33. Sine and cosine of the sum of two angles 63 34. Sine and cosine of the difference of two angles 66 35. Tangent and cotangent of the sum and of the difference of two angles 68 36. Functions of twice an angle in terms of the functions of the angle . . 69 37. Functions of multiple angles . . . . 70 38. Functions of an angle in terms of functions of half the angle ... 72 39. Functions of half an angle in terms of the cosine of the angle ... 72 40. Sums and differences of functions .... 73 41. Trigonometric identities 75 CHAPTER V GENERAL VALUES OF ANGLES. INVERSE TRIGONOMETRIC FUNCTIONS. TRIGONOMETRIC EQUATIONS 42. General value of an angle . . ... ... . . 79 43. General value for all angles having the same sine or the same cosecant 79 44. General value for all angles having the same cosine or the same secant 81 45. General value for all angles having the same tangent or the same co- tangent . . 82 46. Inverse trigonometric functions ... . . 84 47. Trigonometric equations . . . .89 48. General directions for solving a trigonometric equation 90 CHAPTER VI GRAPHICAL REPRESENTATION OF TRIGONOMETRIC FUNCTIONS 49. Variables . . .... ... 50. Constants 51. Functions 52. Graphs of functions 53. Graphs of the trigonometric functions . . 54. Periodicity of the trigonometric functions . 93 93 93 93 95 97 65. Graphs of the trigonometric functions plotted by means of the unit circle 97 CONTENTS ii CHAPTER VII SOLUTION OF OBLIQUE TRIANGLES Section Page 56. Relations between the sides and angles of a triangle 101 57. Law of sines . 102 58. The ambiguous case 104 59. Law of cosines 108 60. Law of tangents . ... ... Ill 61. Trigonometric functions of the half angles of a triangle . ... 113 62. Formulas for finding the area of an oblique triangle 117 CHAPTER VIII THEORY AND USE OF LOGARITHMS 63. Need of logarithms in Trigonometry 119 64. Properties of logarithms 121 65. Common system of logarithms . . 124 66. Rules for determining the characteristic of a logarithm 125 67. Tables of logarithms 128 68. To find the logarithms of numbers from Table I 129 69. To find the number corresponding to a given logarithm .... 133 70. The use of logarithms in computations . . 135 71. Cologarithms 137 72. Change of base in logarithms . . ... . 138 73. Exponential equations ... 140 74. Use of the tables of logarithms of the trigonometric functions . . . 141 75. Use of Table II, angle in degrees and minutes . 142 76. To find the logarithm of a function of an angle ... ... 143 77. To find the acute angle corresponding to a given logarithm . . . 144 78. Use of Table III, angle in degrees and the decimal part of a degree 147 79. To find the logarithm of a function of an angle 148 80. To find the acute angle corresponding to a given logarithm .... 149 81. Use of logarithms in the. solution of right triangles 152 82. Use of logarithms in the solution of oblique triangles 158 Case I. When two angles and a side are given . . .... 158 Case II. When two sides and the angle opposite one of them are given (ambiguous case) 161 Case III. When two sides and the included angle are given . . . 164 Case IV. When all three sides are given 167 83. Use of logarithms in finding the area of an oblique triangle .... 170 84. Measurement of land areas 172 85. Parallel sailing 173 86. Plane sailing 174 87. Middle latitude sailing 175 CONTENTS CHAPTER IX ACUTE ANGLES NEAR 0° OB 90° Section Pare 88. Limits of and as x approaches the limit zero .... 178 x x 89. Functions of positive acute angles near 0° and 90° 179 90. Rule for finding the functions of acute angles near 0° 180 91. Rule for finding the functions of acute angles near 90° 181 92. Rules for rinding the logarithms of the functions of angles near 0° and 90° 182 93. Consistent measurements and calculations 183 CHAPTER X RECAPITULATION OF FORMULAS List of formulas in Plane Trigonometry . . ... . . 189-191 SPHERICAL TRIGONOMETRY CHAPTER I EIGHT SPHERICAL TRIANGLES 1. Correspondence between the parts of a triedral angle and the parts of a spherical triangle . . ... 193 2. Properties of spherical triangles .... 194 3. Formulas relating to right spherical triangles 196 4. Napier's rules of circular parts . . . . 199 5. Solution of right spherical triangles ... . 200 6. The ambiguous case. Two solutions " 203 7. Solution of isosceles and quadrantal triangles 204 CHAPTER II OBLIQUE SPHERICAL TRIANGLES 8. Fundamental formulas ... . 206 9. Law of sines 206 10. Law of cosines 207 11. Principle of Duality 208 12. Trigonometric functions of half the supplements of the angles of a spherical triangle in terms of its sides 210 13. Trigonometric functions of the half sides of a spherical triangle in terms of the supplements of the angles 214 CONTENTS xi Section Pagb 14. Napier's analogies . 215 15. Solution of oblique spherical triangles . 216 16. Case I. (a) Given the three sides ... . 217 17. Case I. (6) Given three angles ... 218 18. Case II. (a) Given two sides and their included angle . . 219 19. Case II. (b) Given two angles and their included side 222 20. Case III. (a) Given two sides and the angle opposite one of them (ambiguous case) 224 21. Case III. (6) Given two angles andtHe side opposite one of them (ambiguous case) 226 22. Length of an arc of a circle in linear units 228 23. Area of a spherical triangle ... . 229 CHAPTER III APPLICATIONS OF SPHERICAL TRIGONOMETRY TO THE CELESTIAL AND TERRESTRIAL SPHERES 24. Geographical terms . . 231 25. Distances between points on the surface of the earth ... . . 232 26. Astronomical problems ... . . 235 27. The celestial sphere 235 28. Spherical coordinates . 237 29. The horizon and meridian system . . 238 30. The equator and meridian system . 240 31. Practical applications 241 32. Relation between the observer's latitude and the altitude of the celes- tial pole 242 33. To determine the latitude of a place on the surface of the earth . . 242 34. To determine the time of day 247 36. To find the time of sunrise or sunset 250 36. To determine the longitude of a place on the earth 250 37. The ecliptic and the equinoxes .... 253 38. The equator and hour circle of vernal equinox system 253 39. The system having for reference circles the ecliptic and the great circle passing through the pole of the ecliptic and the vernal equinox . 256 40. The astronomical triangle 258 41. Errors arising in the measurement of physical quantities .... 259 CHAPTER IV RECAPITULATION OP FORMULAS 42. Right spherical triangles 262 43. Relations between the sides and angles of oblique spherical triangles 262 44. General directions for the solution of oblique spherical triangles . . 264 45. Length of an arc of a circle in linear units 264 46. Area of a spherical triangle 264 PLANE TRIGONOMETRY CHAPTER I TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES SOLUTION OF RIGHT TRIANGLES 1. Trigonometric functions of an acute angle defined. We shall assume that the student is familiar with the notion of the angle between two lines as presented in elementary Plane Geometry. For the present we will confine ourselves to the consideration of acute angles. Let EAD be an angle less than 90°, that is, an acute angle. From B, any point in one of the sides of the angle, draw a perpendicular to the other side, thus forming a right tri- angle, as ABC. Let the capital letters A, B, C denote the angles and the small letters a, b, c the lengths of the corre- sponding opposite sides in the right tri- angle.* We know in a general way from Geometry that the sides and angles of this triangle are mutually dependent. Trigonometry begins by showing the exact nature of this dependence, and for this purpose employs the ratios of the sides. These ratios are called trigonometric functions. The six trigonometric functions of any acute angle, as A, are denoted as follows : sin A, read "sine of A"; COS, A, read "cosine of A"; tan A, read "tangent of A"; esc A, read "cosecant of A"; sec A, read " secant of A " ; cot A, read "cotangent of A." * Unless otherwise stated the hypotenuse of a right triangle will always be denoted by c and the right angle by C. 1 PLANE TEIGONOMETKY These trigonometric functions (ratios) are defined as follows (see figure) : hypotenuse / c opposite side / a (1) sin^=-p— =- s ' hypotenuse \ c adjacent side / b\ (2) cosA = — - (= - ; v ' hypotenuse \ c j (4) esc A = opposite side V a m ccci- h yp° tenuse / - c ^ ' adjacent side \ b opposite side / a\ ,„ adjacent side / b (3) tanA = -^ — — - = r ! (6) cot A = J — — = - v ' adjacent side \ bj ' opposite side \ a The essential fact that the numerical value of any one of these functions depends upon the magnitude only of the angle A, that is, is independent of the point B from which the perpendicular upon < the other side is let fall, is easily established.* These functions (ratios) are of fundamental importance in the study of Trigonometry. In fact, no progress in the subject is pos- sible without a thorough knowledge of the above six definitions. They are easy to memorize if the student will notice that the three in the first column are reciprocals respectively of those directly opposite in the second column. For, a 1 1 sin A = — = - = ; c o esc A a A h 1 1 ■ cos A = - = - = ; c c sec A A ° 1 1 • csc A = - = - = — — - ; a a sin A s6c A = - = -? = ; b b cos A tan^ = t = 7 = — TT b b cot A a cot A = - = - = a a tan^4 * For, let B' he any other point in AD, and B" any point in AE. Draw the perpendicu- lars B'C and B"C" to AE and AD respectively. The three triangles ABC, AB'C, AB"C'\ are mutually equiangular since they are right- angled and have a common angle at A. Therefore they are similar, and we have BC = B'C' = B"C" AB AB' AB" But each of these ratios defines the sine of A. In the same manner we may prove this property for each of the other functions. This shows that the size of the right triangle we choose is imma- terial; it is only the relative and not the actual lengths of the sides of the triangle that are of importance. The student should also note that every one of these six ratios will change in value when the angle A changes in size. FUNCTIONS OF ACUTE ANGLES If we apply the definitions (1) to (6) inclusive to the acute angle B, there results sin B = - ; c cos B = — ; c tan B = - ; a CSC B = P sec B = cots = a Comparing these with the functions of the angle A, we see that sin A = cos B ; cos A = sin B ; tan vl = cot B ; esc ^4. = sec B ; sec A = esc £ ; cot A = tanB. Since A + B = 90° (i.e. 4 and B are complementary) the above results may be stated in compact form as follows : Theorem. A function of an acute angle is equal to the co-function * of its complementary acute angle. Ex. 1. Calculate the functions of the angle A in the right triangle where o = 3, 6 = 4. Solution, c = Va 2 + 6 2 = VsT+16 = V25 = 6. ' Applying (1) to (6) inclusive (p. 2), sin A = | ; esc A = % ; cos A = | ; sec A = | -, tan.4 = | ; cot A = J . Also find all functions of the angle B, and com- A pare results. 6=4 Ex. 2. Calculate the functions of the angle B in the right triangle where a = 3, c = 4. Solution, b = Vc*~^a? = Vl6 - 9 = V7. sin B = V7 4 ' 6=V7 cos B = - ; 4 tan2? = ; esc B = — — ; V7 4 sec B = - ; 3' cotJS = V7 Also find all functions of the angle A, and compare results. * Sine and cosine are called co-functions of each other. Similarly tangent and cotangent, also secant and cosecant, are co-functions. 4 PLANE TRIGONOMETRY Ex. 3. Calculate the functions of the angle A in the right triangle where a = 2 mn, b = m 2 — re- solution. c = Va 2 + 6 2 = Vi m% 2 + m* - 2 m 2 n 2 + n 4 = Vm 4 + 2 to 2 k 2 + n* = m 2 + n 2 . 6=m 2 -n 2 sin .4. = cos^l = tan^l = 2 mn m 2 + n 2 ' m 2 — n 2 m 2 + n 2 ' 2mn esc A = seoA = cot A = ?n 2 + n 2 2 mn m 2 + n 2 m 2 — ri> m 2 - n 2 2 ran m 2 — n 2 Ex. 4. In a right triangle we have given sin .4. = £ and a = 80 ; find c. Solution. From (1), p. 2, we have the formula smA = -- Substituting the values of sin A and a that are given, there results 4_80 5~"c" ; and solving, c = 100. Ans. 2. Functions of 45°, 30°, 60°. These angles occur very frequently in problems that are usually solved by trigonometric methods. It is therefore important to find the values of the trigonometric functions of these angles and to memorize the results. (a) To find the functions of 45°. Draw an isosceles right triangle, as ABC. This makes angle A = angle B = 45°. 6=i Since the relative and not the actual lengths of the sides are of importance, we may assign any lengths we please to the sides satis- fying the condition that the right triangle shall be isosceles. Let us choose the lengths of the short sides as unity, i.e. let a = 1 and 6 = 1. Then c = Va 2 + 6 2 = V2, and we get sin 45° = -^=; csc45° = V2; V2 cos 45° = -4=; sec 45° = V2; V2 tan 45° = 1 ; cot 45° = 1. FUNCTIONS OF ACUTE ANGLES (6) To find the functions of 30° and 60°. Draw an equilateral triangle, as ABD. Drop the perpendicular BC from B to AD, and consider the triangle ABC, where angle A = 60° and angle ABC = 30°. Again take the smallest side as unity, i.e. let 6 = 1. This makes c = AB = AD = 2A C = 2 b = 2, and a = Vc 2 — 6 2 = V4 - 1 = Vs. Therefore sin 60° = ~; csc60° = -?=; 2 V3 cos 60° = see 60° = 2 : tan 60° = V3 ; cot 60° = -==• V3 Similarly, from the same triangle, sin 30° = - ; esc 30° = 2 ; V3 cos 30° = -^ ; sec 30° = — =; V3 tan 30° = ^=; cot 30° = V3. V3' Writing the more important of these results in tabulated form,* we have Angle 30° 45° 60° sin ±=.50 2 V2 ^=.86 + 2 cos ^=.86 + 2 V2 ± = .50 2 tan ^ = .57 + V3 1 V3 = 1.73 + The cosecant, secant, and cotangent are easily remembered as being the reciprocals of the sine, cosine, and tangent respectively. * To aid the memory we observe that the numbers in the first (or sine) row are respec- tively VI, V2, Vi ; each divided by 2. The second (or cosine) row is formed by reversing the order in the first row. The last (or tangent) row is formed by dividing the numbers in the first row by the respective numbers in the second row. PLANE TRIGONOMETRY The student should become very familiar with the 45° right tri- angle and the 30°, 60° right triangle. Instead of memorizing the above table we may then get the values of the functions directly from a mental picture of these right triangles. Ex. 5. Given a right triangle where A = 60°, a = 100 ; find c. Solution. Since we know A (and therefore also any function of A), and the sine of A involves a, which is known, and c, which is wanted, we can find c by using the formula sinA = *. by(l),p. 2 V3 Substituting o = 100, and sin A = sin 60° = — from the above table, we have V3 100 Clearing of fractions and solving for c, we get 200 200 V3 1-7 + ; 117.6 + . Ans. What is the value of B ? Following the method illustrated above, show that 6 = 58.8 + . EXAMPLES Only right triangles are referred to in the following examples. 1. Calculate all the functions of the angle A, having given a = 8, b = 15. Ans. &inA = ~fj, cosA = \fy, tanjl = T 8 j, etc. 2. Calculate the functions of the angle B, having given a = 5, c = 7. . . . V5 _ 5 . „ V24 Ans. sm B = , cos B = - , tan B = , etc. 7 7 5 3. Calculate the functions of the angle A, having given 6 = 2, c =vTl. a rr 2 V7 Ans. \—, — =, — -, etc. \11 Vn 2 4. Calculate the functions of the angle B, having given a = 40, c = 41. Ans - ? 9 t. It. i% etc. 5. Calculate the functions of the angle A, having given a — p, 6 = q. P 9 P Ans. Vp 2 + 32 Vp 2 + q 2 1 , etc. 6. Calculate the functions of the angle A, having given a = Vm 2 + mn, c = m + n. Vm 2 + mn Vmn + n 2 [m Ans. , , -»/— , etc. m + n m + n V n 7. Calculate the functions of the angle B, having given a = Vm^+li 2 " c = m + n. V2mn Vm 2 + n 2 / 2«n Ans. — , , -./_- etc. m + n m + n \m 2 + n 2 FUNCTIONS OF ACUTE ANGLES 7 8. Given sin.4 = f, c = 200.5; calculate a. Ans. 120.3. 9. Given cosA = .44, c = 30.5; calculate ft. Ans. 13.42. 10. Given tan A = 1 j-, & = ff; calculate c. Ans. T 9 T Vl30. 11. Given A = 30°, a = 25; calculate c. Also find B and b. Ans. c = 50, B = 60°, 6 = 25 V& 12. Given B = 30°, c = 48 ; calculate 6. Also find A and a. Ans. 5 = 24, A = 60°, a = 24 V3. 13. Given £ = 45°, 6 = 20 ; calculate c. Also find A and a. ^ms. c = 20 V2, 4 = 45°, a = 20. 3. Solution of right triangles. A triangle is composed of six parts, three sides and three angles. To solve a triangle is to find the parts not given. A triangle can be solved if three parts, at least one of which is a side, are given.* A right triangle has one angle, the right angle, always given ; hence a right triangle can be solved if two sides, or one side and an acute angle, are given. One of the most impor- tant applications of Trigonometry f is the solution of triangles, and we shall now take up the solution of right triangles. The student may have noticed that Examples 11, 12, 13, of the last section were really problems on solving right triangles. When beginning the study of Trigonometry it is important that the student should draw the figures connected with the problems as accurately as possible. This not only leads to a better understanding of the problems themselves, but also gives a clearer insight into the 'meaning of the trigonometric functions and makes it possible to test roughly the accuracy of the results obtained. For this purpose the only instruments necessary are a graduated ruler and a protractor. A protractor is an instrument for measuring angles. On the inside of the back cover of this book will be found a Granville's Transparent Combined Ruler and Protractor, with directions for use. The ruler is graduated to inches and centimeters and the protractor to degrees. The student is advised to make free use of this instrument. 4. General directions for solving right triangles. First step. Draw a figure as accurately as possible representing the triangle in question. Second step. When one acute angle is known, subtract it from 90° to get the other acute angle. * It is assumed that the given conditions are consistent, that is, that it is possible to construct the triangle from the given parts. t The name Trigonometry is derived from two Greek "words which taken together mean " I measure .a triangle." 8 PLANE TRIGONOMETRY Third step. To find an unknown part, select from, (1 ) to (6), p. 2, a formula involving the unknown pari and two known parts, and then solve for the unknown part. Fourth step. Check the values found by noting whether they satisfy relations different from those already employed in the third step. A. convenient numerical check is the relation, a 2 = c 2 - b 2 = (c + b) (c - b). Large errors may be detected by measurement. Since the two perpendicular sides of a right triangle may be taken as base and altitude, we have at once Area of a right triangle = %r- • In the last section the functions 30°, 45°, 60°, were found. In more advanced treatises it is shown how to calculate the functions of angles in general. We will anticipate some of these results by making use of the following table where the values* of the trigonometric functions for each .degree from 0° to 90° inclusive are correctly given to four or five significant figures. In looking up the function of an angle between 0° and 45° inclu- sive, we look for the angle in the extreme left-hand vertical column. The required value of the function will be found on the same hori- zontal line with the angle, and in the vertical column having that function for a caption at the top. Thus, sin 15° = .2588, cot 41° = 1.1504, etc. Similarly, when looking up the function of an angle between 45° and 90° inclusive we look in the extreme right-hand vertical column. The required value of the function will be found on the same hori- zontal line with the angle as before, but in the vertical column hav- ing that function for a caption at the bottom. Thus, cos 64° = .4384, sec 85° = 11.474, etc. When we have given the numerical value of the function of an angle, and wish to find the angle itself, we look for the given num- ber in the columns having the given function as a caption at the top * Also called the natural values of the trigonometric functions in contradistinction to their logarithms (see Tables II and III of Granville's Four-Place Tables of Logarithms). FUNCTIONS OF ACUTE ANGLES Table A NATURAL VALUES OF THE TRIGONOMETRIC FUNCTIONS Angle sin COS tan cot sec CSC 0° .0000 1.0000 .0000 CO 1.0000 CO 90° 1° .0175 .9998 .0175 57.290 1.0002 57.299 89° 2° .0349 .9994 .0349 28.636 1.0006 2S.654 88° 3° .0523 .9986 .0524 19.081 1.0014 19.107 87° 4° .0698 .9976 .0699 14.301 1.0024 14.336 86° 5° .0872 .9962 .0875 11.430 1.0038 11.474 85° 6° .1045 .9945 .1051 9.5144 1.0055 9.5668 84° 7° .1219 .9925 .1228 8.1443 1.0075 8.2055 83° S° .1392 .9903 .1405 7.1154 1.0098 7.1853 82° 9° .1564 .9877 .1584 6.3138 1.0125 6.3925 81° 10° .1736 .9848 .1763 5.6713 1.0154 5.75SS 80° 11° .1908 .9816 .1944 5.1446 1.0187 5.2408 79° 12° .2079 .9781 .2126 4.7046 1.0223 4.8097 78° 13° .2250 .9744 .2309 4.3315 1.0263 4.4454 77° 14° .2419 .9703 .2493 4.0108 1.0306 4.1336 76° 15° .2588 .9659 .2679 3.7321 1.0353 3.8637 75° 16° .2756 .9613 .2S67 3.4874 1.0403 3.6280 74° 17° .2924 .9563 .3057 3.2709 1.0457 3.4203 73° 18° .3090 .9511 .3249 3.0777 1.0515 3.2361 72° 19° .3256 .9455 .3443 2.9042 1.0576 3.0716 71° 20° .3420 .9397 .3640 2.7475 1.0642 2.9238 70° 21° .3584 .9336 .3839 2.6051 1.0711 2.7904 69° 22° .3746 .9272 .4040 2.4751 1.0785 2.6695 68° 23° .3907 .9205 .4245 2.3559 1.0864 2.5593 67° 24° .4067 .9135 .4452 2.2460 1.0946 2.4586 66° 25° .4226 .9063 .4663 2.1445 1.1034 2.3662 65° 26° .4384 .8988 .4877 2.0503 1.1126 2.2812 64° 27° .4540 .8910 .5095 1.9626 1.1223 2.2027 63° 28° .4695 .8829 .5317 1.8807 1.1326 2.1301 62° 29° .4848 .8746 .5543 1.8040 1.1434 2.0627 61° 30° .5000 .8660 .5774 1.7321 1.1547 2.0000 60° 31° .5150 .S572 .6009 1.6643 1.1666 1.9416 59° 32° .5299 .8480 .6249 1.6003 1.1792 1.8871 58° 33° .5446 .8387 .6494 1.5399 1.1924 1.8361 57° 34° .5592 .8290 .6745 1.4826 1.2062 1.7883 56° 35° .5736 .8192 .7002 1.4281 1.2208 1.7434 55° 36° .5878 .8090 .7265 1.3764 1.2361 1.7013 54° 37° .6018 .7986 .7536 1.3270 1.2521 1.6616 53° 38° .6157 .7880 .7813 1.2799 1.2690 1.6243 52° 39° .6293 .7771 .8098 1.2349 1.2868 1.5890 51° 40° .6428 .7660 .8391 1.1918 1.3054 1.5557 50° 41° .6561 .7547 .8693 1.1504 1.3250 1.5243 49° 42° .6691 .7431 .9004 1.1106 1.3456 1.4945 48° 43° .6820 .7314 .9325 1.0724 1.3673 1.4663 47° 44° .6947 .7193 -.9657 1.0355 1.3902 1.4396 46° 45° .7071 .707] 1.0000 1.0000 1.4142 1.4142 45° cos sin cot tan CSC see Angle 10 PLANE TRIGONOMETRY or bottom. If we find it in the column having the given function as a top caption, the required angle will be found on the same horizontal line and in the extreme left-hand column. If the given function is a bottom caption, the required angle will be found in the extreme right-hand column. Thus, let us find the angle x, having given tan x = .7536. In the column with tan as top caption we find .7536. On the same horizontal line with it, and in the extreme left-hand column, we find the angle x = 37°. Again, let us find the angle x, having given sin a; = .9816. In the column with sin as bottom caption we find .9816. On the same horizontal line with it, and in the extreme right-hand column, we find the angle x = 79°. The following examples will further illustrate the use of the table. Ex. 1. Given A = 36°, c = 2W ; solve the right triangle. Also find its area. Solution. First step. Draw a figure of the triangle indicating the known and unknown parts. &=? Second step. B = 90°-A = 90°- 35°= 56°. Third step. To find a use formula (1), p. 2, namely, a sin A = -■ c Substituting the value of sin A = sin 35° = .5736 (found from the table) and c = 267, we have „ .6736 = —. 267 Solving for a, we get a = 153.1.* » Multiplying, sin 36° = .5736 267 40152 34416 11472 a =153.1512 Since oar table gives not more than the first four significant figures of the sine of an angle, it follows, in general, that all but the first four significant figures of the product are doubtful. The last three figures of the above product should therefore be omitted, for the result will not be more accurate if they are retained. To illustrate this in the above example, suppose we take the sine of 35° from a five-place table, that is, a table which gives the first live significant figures of the sine. Then sin 35° = .57358 267 401506 344148 114716 0=153.14586 Comparing, we see that the two values of a agree in the first four significant figures only. Hence we take a = 153.1. FUNCTIONS OF ACUTE ANGLES To find 6 use formula (2), p. 2, namely, 11 cos .4 = Substituting as before, we have .8192 = — , 267 since from the table cos A = cos 35° = .8192. Hence 6 = 218.7. Fourth step. By measurements we now check the results to see that there are no large errors. As a numerical check we find that the values of a, 6, c sat- isfy the condition c 2 = a 2 + &2. To find the area of the triangle, we have ab 153.1x218.7 ,.,,., Area = — = = 16,741. 2 2 ' Ex. 2. A ladder 30 ft. long leans against the side of a building, its foot being 15 ft. from the building. What angle does the ladder make with the grc jnd ? Solution. Our figure shows a right triangle with hypotenuse and side adjacent to the re- quired angle (= x) given. Hence cosx = ^ = £=.5 = .5000. This number is found in the column having cos at the bottom and opposite 60°. Hence x = 60°- Ans. We shall now derive three formulas by means of which the work of solving right triangles may be simplified. From (1), (2), (3), p. 2, a sin A = -i or, c a = o sin A ; . b cos A = - > or, o b = e cos A ; tan A = - 1 or, a = bt&nA. These results may be stated as follows : (7) Side opposite an acute angle = hypotenuse x sine of the angle. (8) Side adjacent an acute angle = hypotenuse x cosine of the angle. (9) Side opposite an acute angle = adjacent side x tangent of the angle. 12 PLANE TEIGONOMETKY EXAMPLES Solve the following right triangles (C = 90°). No. Given Parts Required Parts Area 1 ^ = 60° 6 = 4 5 = 30° c = 8 a = 6.928 13.856 2 4 = 30° a = 3 £ = 60° c = 6 6 = 5.196 7.794 3 a = 6 c= 12 ^1 = 30° 5 = 60° 6= 10.39 31.18 4 a = 4 6 = 4 ^1 = 45° £ = 45° c = 5.657 8 5 a = 2 c = 2.8284 A = 45° £ = 45° 6 = 2 2 6 a = 51.303 c= 150 4 = 20° £ = 70° 6 = 140.95 3615.6 7 5 = 51° c = 250 A = 39° a = 157.3 5 = 194.3 15282 8 A = Z6° c= 1 5 = 54° a= .5878 6 = .809 .2378 9 c = 43 a = 38.313 A= 63° £ = 27° 6 = 19.52 373.9 10 6 = 9.696 e = 20 ^ = 61° £ = 29° a = 17.492 84.8 11 a = 137.664 c = 240 ^4 = 35° £ = 55° 6 = 196.6 13532 12 ^1 = 75° a = 80 5 = 15° 6 = 21.43 c = 82.82 857 13 ;1 = 25° a = 30 £ = 65° 6 = 64.336 c = 70.99 965 14 £ = 55° 6 = 10 4 = 35° a = 7.002 c = 12.208 35 15 £ = 15° 6 = 20 A = 75° a = 74.64 c = 77.28 746.5 16 a = 36.4 6= 100 ^1 = 20° £ = 70° c = 106.4 1820 17 a = 23.315 6 = 50 ^1 = 25° £ = 65° c= 55.17 583 18 a= 17.1 c = 50 ^1 = 20° £ = 70° 6 = 46.985 402 19 4= 10° 6 = 30 5=80° a = 5.289 c = 30.46 79 20 .4 = 20° c = 80 £ = 70°" a = 27.36 6 = 75.176 1028 21 £ = 86° 6= .08 .4 = 4° a = .00559 c = .0802 .0002 22 £ = 32° c = 1760 A = 58° 6 = 932.62 a= 1492.5 696968 23 a = 30.21 c = 33.33 4 = 65° £ = 25° 6 = 14.085 213 24 a = 13.395 6 = 50 A = 15° £ = 75° c = 51.77 335 25 6 = 93.97 c= 100 ^1 = 20° £ = 70° a = 34.2 1607 26. A tree is broken by the wind so that its two parts form with the ground a right-angled triangle. The upper part makes an angle of 35° with the ground, and the distance on the ground from the trunk to the top of the tree is 50 ft. Find the length of the tree. Ans. 96.05 ft. 27. In order to find the breadth of a river, a dis- tance AB was measured along the bank, the point A being directly opposite a tree C on the other side. If the angle ABC was observed to be 55° and AB 100 ft., find the breadth of the river. Ans. 142.8 ft. 28. Two forts defending a harbor are 2 mi. apart. From one a, hostile battleship is observed due south and from the other 15° east of south. How far is the battleship from the nearest fort ? Ans. 7.464 mi. 29. A vessel whose masts are known to reach 100 ft. above her water line subtends in a vertical plane an angle of 5° to an observer in a rowboat. How far is the boat from the vessel ? Ans. 1143 ft. FUNCTIONS OF ACUTE ANGLES 13 30. The vertical central pole of a circular tent is 20 ft. high, and its top is fastened by ropes 40 ft. long to stakes set in the ground. How far are the stakes from the foot of the pole, and what is the inclination of the ropes to the ground? Ans. 34.6ft; 30°. 31. A wedge measures 10 in. along the side and the angle at the vertex is 20°. Find the width of the base. Ans. 3.47 in. 32. At two points A, B, 400 yd. apart on a straight horizontal road, the summit of a hill is observed ; at A it is due north with an elevation of 40°, and at B it is due west with an elevation of 27°. Find the height of the hill. Ans. 522.6 ft. 5. Solution of isosceles triangles. An isosceles triangle is divided by the perpendicular from the vertex to the base into two equal right triangles ; hence the solution of an isosceles triangle can be made to depend on the solution of one of these right triangles. The following examples will illustrate the method. Ex. 1. The equal sides of an isosceles triangle are each 40 in. long, and the equal angles at the base are each 25°. Solve the triangle and find its area. Solution. B = 180° - (A + C) = 180° - 50° = 130°. Drop the perpendicular BD to AC. B AD = AB cos A = 40 cos 25° by (8), p. 11 w = 40 x .9063 = 36.25. Therefore by (7), p. 11 by (9), p. 11 AC = 2 AD = 72.50 in. To find the area we need in addition the altitude BD. BD = AB smA = 40 sin 25° = 40 x. 4226 = 16. 9. Check : BD = AD tan 25° = 36.25 x .4663 = 16.9. Also, Area = \AC x BD = 612.6 sq. in. Ex. 2. A barn 60 ft. wide has a gable roof whose rafters are 30 V2 ft. long. What is the pitch of the roof, and how far above the eaves is the ridgepole ? Solution. Drop a perpendicular from B to AD. Then AC 30 1 cos X = = = -^B 30 V2 V2 Hence x = 45° = pitch of the roof. Also, BC = 45 sin x by (8), p. 11 1 = 30V2- vS Check : AB = V^ C 2 + JSG 2 = V(30)* + (30)* : = 30 ft. = height of the ridgepole above the eaves. V1800 = 30 V2. 14 . PLANE TKIGONOMETKY EXAMPLES 1. The equal sides of an isosceles triangle are each 12 in. long, and the angle at the vertex is 120°. Find the remaining parts and the area. Ans. Base = 20.78 in.; base angles = 30°; area = 62.35 sq. in. 2. The equal angles of an isosceles triangle are each 35°, and the base is 393.18 in. Find the remaining parts. Ans. Vertex angle = 110° ; equal sides = 240 in. 3. Given the base 300 ft. and altitude 150 ft. of an isosceles triangle ; solve the triangle. Ans. Vertex angle = 90° ; equal angles = 45° ; equal sides = 212. 13 ft. 4. The base of an isosceles triangle is 24 in. long and the vertex angle is 48° ; find the remaining parts and the area. Ans. Equal angles = 06° ; equal sides = 29.5 in. ; area = 323.4 sq. in. 5. Each of the equal sides of an isosceles triangle is 50 ft. and each of its equal angles is 40°. Find the base, the altitude, and the area of the triangle. Ans. Alt. = 32.14 ft. ; base = 76.6 ft. ; area = 1231 sq. ft. 6. The base of an isosceles triangle is 68.4 ft. and each of its equal sides is 100 ft. Find the angles, the height, and the area. Ans. 40°, 70°; 93.97 ft.; 3213.8 sq.ft. 7. The base of an isosceles triangle is 100 ft. and its height is 35.01 ft. Find its equal sides and the angles. Ans. 61.04 ft.; 35°, 110°. 8. The base of an isosceles triangle is 100 ft. and the equal angles are each 65°. Find the equal sides, the height, and the area. Ans. 118.3 ft; 107.23 ft; 5361.5 sq.ft. 9. The ground plan of a barn measures 40 x 80 ft. and the pitch of the roof is 45° ; find the length of the rafters and the area of the v? hole roof, the horizontal projection of the cornice being 1 ft. Ans. 29.7 ft. ; 4870.5 sq. ft. 6. Solution of regular polygons. Lines drawn from the center of a regular polygon of n sides to the vertices are the radii of the circum- scribed circle and divide the poly- gon into n equal, isosceles triangles. The perpendiculars from the center to the sides of the polygon are the radii of the inscribed circle and divide these n equal isosceles tri- angles into 2 n equal right triangles. Hence the solution of a regular polygon depends on the solution of one of these right triangles. FUNCTIONS OF ACUTE ANGLES 15 From Geometry we know that the central angle ABC ■ in the right triangle ABD the 180° 360° n ; hence Also, angle x = AD = - = half the length of one side, AB = R = radius of circumscribed circle, BD = r = radius of inscribed circle, p = no = perimeter of polygon, pr ~ = area of polygon. EXAMPLES 1. One side of a regular decagon is 10 in. ; find radii of inscribed and circum- scribed circles and area of polygon. Solution. Since n = 10, in this example we have x = - 180° _ 180° "IF = 18°. Then R = ■ and Check Also, hence r = - sin 18° 5 .3090 5 = 16.18 in., = 15.39 in. tan 18° .3249 : r = R cos 18° = 16.18 x .9511 = 15.39. p = 10 x 10 = 100 in. = perimeter of polygon ; pr _ 100 x 15.39 2 ~ 2 = 769. 5 sq. in. = area. 2. The side of a regular pentagon is 24 ft. ; find R, r, and area. Ana. 20.42 ft.; 16.52 ft. ; 991.2 sq. ft. 3. Find the remaining parts of a regular polygon, having given (a) n = 9, c = 12. Arts. R = 17.54 ; r = 16.48 ; area = 889.9. (b) n = 18, R = 10. r = 9.848 ; c = 3.472 ; area = 307.7. (c) n = 20, R = 20. r = 19.75 ; c = 6.256 ; area = 1236. (d)n=12, r=8. iJ = 8.28; c = 4.29; area = 206. 4. The side of a regular hexagon is 24 ft. Find the radii of the inscribed and circumscribed circles ; also find the difference between the areas of the hexagon and the inscribed circle, and the difference between the areas of the hexagon and the circumscribed circle. Ans. R = 24 ft.; r = 20.8 ft.; 138.4 sq. ft.; 312 sq. ft. 5. If c be the side of a regular polygon of n sides, show that x, 1 180 ° A R = -c esc and r ■■ 2 n 1 x 180° - c cot 2 71 16 PLANE TKIGONOMETKY 6. If r be the radius of a circle, show that the side of the regular inscribed 180° polygon of n sides is 2 r sin , and that the side of the regular circumscribed • „ ♦ 180 ° n polygon is 2 r tan 7. Interpolation. In the examples given so far we have needed the functions of such angles only as were explicitly given in our table ; that is, the number of degrees in the angle involved was given by a whole number. It is evident that such will not always be the case. In general, our problems will involve angles expressed in degrees and parts of a degree, as 28.4°, 5.63°, 10° 13', 72° 27.4', 42° 51' 16", etc. In order to find from the table the numerical value of the function of such an angle not given in the table, or to find the angle corre- sponding to a given numerical value of some function not found in the table, we use a process called interpolation. This is based on the assumption that a change in the angle causes a proportional change in the value of each function, and conversely, provided these changes are small* To illustrate ; from the table we have sin 38° = .6157 sin 37° = .6018 Subtracting, .01 39 = difference for one degree ; that is, at 37° a change of one degree in the angle causes a change in the value of the sine of .0139. If, then, x is any other small change in the angle from 37°, and d the corresponding change in the value of the sine, we must have, near 37°, V:x\: .0139 : d, .-. d = .0139x, if x is expressed in the decimal parts of a degree. For example, let us tabulate the values of the sines of all angles from 37° to 38° at intervals of 0.1 of a degree. . . sin 37.1° = .6018 + .0014 = .6032 .-. sin 37.2° = .6018 + .0028 = .6046 sin 37.3° = .6018 + .0042 = .6060 .-. sin 37.4° = .6018 + .0056 = .6074 sin 37.5° = .6018 + .0070 = .6088 .-. sin37.6° = .0018 + .0083 = .6101 ,-. sin 37.7° = .6018 + .0097 = .6115 sin 37.8° = .6018 + .0111 = .6129 .-. sin 37.9° = .6018 + .0125 = .6143 * This condition is most important. The change in value of the cotangent for one degree is very large when the angle is very small. In this ease the table -would therefore lead to very inaccurate results if interpolation -was used for cotangents of small angles (see Chapter IX, p. 178). X d 0.1° .0014 0.2° .0028 0.3° .0042 0.4° .0056 0.5° .0070 0.6° .0083 0.7° .0097 0.8° .0111 0.9° .0125 FUNCTIONS OP ACUTE ANGLES 17 The following examples will further illustrate the process of interpolating. (a) To find the function of a given angle when the angle is not found in the table. Ex. 1. Find sin 32.8°. Solution. The sine of 32.8° must lie between sin 32° and sin 33°. Prom the table on p. 9, sin 33° = .5446 sin 32° = .5299 .0147 = difference in the sine (called the tabular difference) corresponding to a difference of 1° in the angle. Now in order to find sin 32.8°, we must find the difference in the sine corresponding to .8° and add it to sin 32°, for the sine will be increased by just so much when the angle is increased from 32° to 32.8°. Denoting by d the difference corre- sponding to .8°, we have 1° : .8° : : .0147 : d, or, d = .0118. Hence .sin 32° = . 5299 d - .0118 = difference for .8° sin 32.8° = .5417. Ans. Ex. 2. Eind tan 47° 25'. Solution. The tangent of 47° 25' must lie between tan 47° and tan 48°. Erom the table, tan48° = l.llQ6 tan 47° = 1.0724 .0382 = tabular difference correspond- ing to a difference of 60' (=1°) in the angle. Denoting by d the difference corresponding to 25', we have 60' : 25' : : .0382 : d, or, d = .0159. Hence tan 47° = 1.0724 d = .0159 = difference for 25'. .-. tan 47° 25' = 1.0883. Ans. Ex. 3. Eind cos 68. 57°. Solution. The cosine of 68. 57° must lie between cos 68° and cos 69°. From the table, cos 68° = .3746 cos 69° = .3584 .0162 = tabular difference correspond- ing to a difference of 1° in the angle. Denoting by d the difference corresponding to. 57°, we have l°:.57°::.0162:d, or, d = .0092. 18 PLANE TRIGONOMETRY Since the cosine decreases as the angle increases, this difference must be subtracted* from cos 68° in order to get cos 68. 57° Hence cos 68° = .3746 d = .0092 = difference for. 57°. .-. cos 68. 57° = .3654. Ans. (S) To find an angle when the given numerical value of a function of the angle is not found in the table. Ex. 4. Find the angle whose tangent is .4320. Solution. This problem may also be stated : Having given tan x = .4320, to find the angle x. We first look up and down the columns with tan at top or bottom, until we find two numbers between which .4320 lies. These are found to be .4245 and .4452, the former being tan 23° and the latter tan 24°. We then know that the required angle x must lie between 23° and 24°. To find how far ( = y) beyond 23° the angle x lies, we first find the difference between tan 23° and tan x ; thus, tan x = . 4320 tan 23° = .4245 .0075 = difference in the tangent corre- sponding to the excess of the angle x over 23° ; denote this excess by y. Also, tan 24° = .4452 tan 23° = .4245 .0207 = tabular difference correspond- ing to a difference of 1° in the angle. Then, as before, 1° :y:\. 0207:. 0075, or, y = .36°. Hence x = 23° + "y = 23.36°. Ans. In case we want the angle expressed in degrees and minutes, we can either multiply .36° by 60, giving 21.6' so that the required angle is 23° 21.6', or else we can find y in minutes at once by using instead the proportion 60' : y : : .0207 : .0075, or, y = 21.6'. Hence x = 23° + y = 23° 21.6'. Ans. EXAMPLES 1. Verify the following : (a) sin 51.6° = .7836. (f) esc 80.3° = 1.0145. (k) sec 25° 2.5' = 1.1038. (b) tan 27.42° = .5188. (g) sin 43° 18' = .6858. (1) esc 72° 54' = 1.0463. (c) cos79.9° = .1753. (h) cos 84° 42' = .0924. (m) sin 58° 36.2' = .8536. (d) cot 65.62° = .4532. (i) tan 31° 7.8' = .6040. (e) sec 12.37° = 1.0238. (j) cot 11° 43.4' =4.8263. * In the case of the sine, tangent, and secant this difference is always added, because these functions increase when the angle increases (the angle being acute). In the case of the cosine, cotangent, and cosecant, however, this difference is always subtracted, because these functions decrease when the angle increases. It is always the function of the smaller of the two angles that this difference is added to or subtracted from. TEEMS IN TRIGONOMETRIC PROBLEMS 19 2. Find the angle x, having given (a) sin x = .5280. (b) tana; = .6344. (c) sec x = 1.2122. (d) cos x = . 9850. (e) cot x= 3.5249. (f) esc x = 1.7500. (g) sin x = . 9425. (h) cos x = . 2118. (i) tanx = 1.1652. (j) cot x = .0803. (k) sec x = 4.6325. (1) esc x = 1.2420. (m) sin x = .7100. (n) cos x = .9999. (o) tan x = . 9845. (p) cot x = 8.6892. Ans. x = 31.87°. x = 32.39°. x = 34.41°. x = 9.93°. x = 15.85°. x = 34.85°. x = 70° 28.96'. x = 77.77°. x = 49° 21.4'. x = 85.41°. x = 77.51°. x = 53.63°. x = 45° 14.3'. x = 0° 30'. x = 44°33'. x = 6°36.1'. 8. Terms occurring in trigonometric problems. The vertical line at a point is the line which coincides with the plumb line through that point. A horizontal line at a point is a line which is perpendicular to the vertical line through that point. A vertical plane at a point is a plane which contains the vertical line through that point. The horizontal plane at a point is the plane which is perpendicu- lar to the vertical line through that point. A vertical angle is one lying in a vertical plane. A horizontal angle is one lying in a horizontal plane. The angle of elevation of an object above the horizontal plane of the observer is the ver- tical angle between the line drawn from the observer's eye to the object, and a hori- zontal line through the eye. The angle of depression of an object below the horizontal plane of the observer is the vertical angle between the line drawn from the observer's eye to the object, and a horizontal line through the eye. The horizontal distance between two points is the distance from one of the two points to the vertical line drawn through the other. Horizontal Line Horizontal Line 20 PLANE TKIGONOMETEY The vertical distance between two points is the distance from one of the two points to the horizontal plane through the other. Thus, let BC be the vertical line at B, and let the horizontal plane at A cut this vertical line in C ; then AC is called the horizontal distance between A and B and BC the vertical distance. The Mariner's Compass is divided into 32 equal parts ; hence each part = 360° -*- 32 = 11£°. The following figure shows how the different divisions are designated. North, south, east, and west are called the cardinal points, and on paper these directions are usually taken as upward, downward, to the right, and to the left respectively. The direction of an object from an observer at C may be given in several ways. Thus, A in the figure is said to bear N.E. by E. from C, or from C the bearing of A is N.E. by E. In the same way the bearing of C from A is S.W. by W. The point A is 3 points north of east and 5 points east of north. Also, E. 33|° N. means the same as N.E. by E. In order to illustrate the application of the trigonometric functions (ratios) to the solution of practical examples, we shall now give a variety of problems on finding heights, dis- ^ ^ tan ces , angles , areas , etc. In solving these prob- lems it is best to follow some definite plan. In general we may proceed as follows : (a) Construct a draw- ing to some convenient scale which will show the relations between the given and the re- quired lines and angles. (b) If necessary draw any auxiliary lines that will aid in the solution, and decide on the simplest steps that will solve the problem. (c) Write down the formulas needed, make the calculations, and check the results. EXAMPLES 21 EXAMPLES Solve the following right triangles (C = 90°). No. Given Parts Required Parts 1 a = 60 c = 100 A = 36° 52' £ = 53° 8' 6 = 80 2 a = 16.98 c = 18.7 A = 65° 14' £ = 24° 46' 6 = 7.833 3 a = 147 c= 184 A =53° 2' £ = 36° 58' 6 = 110.67 4 A = 34° 15' a = 843.2 £ = 55° 45' c = 1498.5 6 = 1238.7 5 4 = 31° 14.2' c = 2.934 £ = 58° 45.8' a = 1.521 6 = 2.509 6 £ = 47.26° c = 4.614 ^ = 42.74° a = 3.131 6 = 3.389 7 A = 23.5° c = 627 £ = 66.5° a = 250 6 = 575 8 A = 28° 5' c =. 2280 £ = 61° 55' a = 1073 6 = 2011 9 .5 = 43.8° 6 = 50.94 A = 46.2° a = 53.13 c = 73.6 10 £ = 6° 12.3' c = 3721 ^1 = 83° 47. 7' a = 3699 6 = 402.2 11 a = .624 c= .91 A = 43° 18' £ = 46° 42' 6 = .6623 12 a= 5 6 = 2 ^. = 68° 12' £ = 21° 48' c = 5.385 13 a = 101 6 = 116 A = 41° 3' £ = 48° 57' c = 153.8 14 ^ = 43.5° c = 11.2 £ = 46.5° a = 7.71 6 = 8.124 15 £ = 68° 50' a = 729.3 4 = 21° 10' 6 = 1884 c = 2020 16 4 = 58.65° c = 35.73 £ = 31.35° a = 30.51 5= 18.59 17 B = 10.85° c = .7264 A = 79.15° a = .7134 6 = .1367 18 a = 24.67 6 = 33.02 A = 36° 46' £ = 53° 14' c = 41.22 19 B = 21° 33' 51" a = .821 A = 68° 26' 9" 6 = .3244 c = .8827 20 ^. = 74°0'18" c = 275.62 £ = 15° 59' 42" a = 264.9 6 = 75.93 21 A = 64° 1.3' 6 = 200.05 £ = 25° 58.7' a = 410.6 c = 456.7 22 6 = .02497 c = .04792 .4 = 58° 36' £ = 31° 24' a = .0409 23 6 = 1.4367 c = 3.4653 A = 65° 30' £ = 24° 30' a = 3.153 24. The length of a kite string is 250 yd., and the angle of elevation of the kite is 40°. Pind the height of the kite, supposing the line of the kite string to be straight. Ans. 160.7 yd. 25. At a point 200 ft. in a horizontal line from the foot of a tower the angle of elevation of the top of the tower is observed to be 60°. Pind the height of the tower. Ans. 346 ft. 26. A stick 10 ft. in length stands vertically on a horizontal plane, and the length of its shadow is 8.391 ft. Pind the angle of elevation of the sun. Ans. 50°. 27. Prom the top of a rock that rises vertically 80 ft. out of the water the angle of depression of a boat is found to be 30° ; find the distance of the boat from the foot of the rock. Ans. 138.57 ft. 28. Two ships leave the same dock at the same time in direc- tions S.W. by S. and S.E. by E. at rates of 9 and 9.5 mi. per hour respectively. Find their distance apart after 1 hr. Ans. 13.1 mi. 22 PLANE TEIGONOMETEY 29. From the top of a tower 120 ft. high the angle of depression of an obj on a, level with the base of the tower is 27° 43'. What is the distance of object from the top and bottom of the tower 1 Ans. 258 ft., 228 30. A ship is sailing due east at the rate of 7.8 mi. an hour. A headlanc observed to bear due north at 10.37 a.m. and 33° west of north at 12.43 p, Find the distance of the headland from each point of observation. Ans. 25.22 mi., 30.08 I 31. A ship is sailing due east at a uniform rate of speed. At 7 a.m. a lig] house is observed bearing due north, 10.32 mi. distant, and at 7.30 a.m. it bei 18° 13' west of north. Find the rate of sailing of the ship and the bearing the lighthouse at 10 a.m. Ans. 6.79 mi. per hour, 63° 8' W. of 32. From the top of a tower the angle of depression of the extremity of horizontal base line 1000 ft. in length, measured from the foot of the tower, observed to be 21° 16' 37"- Find the height of the tower. Ans. 389.5 : 33. The length of the side of a regular octagon is 12 in. Find the radii of t inscribed and circumscribed circles. Ans. 14.49 in., 15.69 i 34. What is the angle of elevation of an inclined plane if it rises 1 ft. in horizontal distance of 40 ft. ? Ans. 1°2I 35. A ship is sailing due N.E. at the rat^ of 10 mi. an hour. Find the rate which she is moving due north. Ans. 7.07 mi. per lion 36. A ladder 40 ft. long may be so placed that it will reach a window 33 i high on one side of the street, and by turning it over without moving its foi it will reach a window 21 ft. high on the other side. Find the breadth of tl street. Ans.' 56.64 f 37. At a point midway between two towers on a horizontal plane the angl of elevation of their tops are 30° and 60° respectively. Show that one tower three times as high as the other. 38. A man in a balloon observes that the bases of two towers, which are mile apart on a horizontal plane, subtend an angle of 70°. If he is exactly abo' the middle point between the towers, find the height of the balloon. Ans. 3770 f 39. In an isosceles triangle each of the equal angles is 27° 8' and each of tl equal sides 3.088. Solve the triangle. Ans. Base = 5.49 40. What is the angle of elevation of a mountain slope which rises 238 ft. i a horizontal distance of one eighth of a mile ? Ans. 19° fit 41. If a chord of 41.36 ft. subtends an arc of 145° 37', what is the radius i the circle ? Ans. 21.65 f 42. If the diameter of a circle is 3268 ft. , find the angle at the center sul tended by an arc whose chord is 1027 ft. Ans. 36° 37.8 43. From each of two stations east and west of each other the angle > elevation of a balloon is observed to be 45°, and its bearings N.W. and N.l respectively. If the stations are 1 mi. apart, find the height of the balloon. Ans. 3733 f EXAMPLES 23 44. In approaching a fort situated on a plain, a reconnoitering party finds at one place that the fort subtends an angle of 10°, and at a place 200 ft. nearer the fort that it subtends an angle of 15°. How high is the fort and what is the distance to it from the second place of observation ? Bint. Denoting the height by y and the distance by x, we have y = x tan 15", by (9), p. 11 also, y = (x + 200) tan 10°. . by (9), p. 11 Solve these two simultaneous equations for x and y, substituting the values of tan 15° and tan 10° from the table on p. 9. Ang % _ ggg t t. r y = 103 ft. 45. A cord is stretched around two wheels with radii of 7 ft. and 1 ft. respec- tively, and with their centers 12 ft. apart. Prove that the length of the cord is 12V3+10:rft. 46. A flagstaff 25 'ft. high stands on the top of a house. From a point on the plain on which the house stands, the angles of elevation of the top and the bottom of the flagstaff are observed to be 60° and 45° respectively. Find the height of the house. Ans. 34.15 ft. 47. A man walking on a straight road ob- serves at one milestone a house in a direction making an angle of 30° with the road, and at the next milestone the angle is 60°. How far is the house from the road ? Ans. 1524 yd. 48. Find the number of square feet of pave- ment required for the shaded portion of the streets shown in the figure, all the streets being 50 ft. wide. OQ7KA Ans. =^ + 7500 = 24094. V3 CHAPTEE II TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 9. Generation of angles. The notion of an angle, as usually pre- sented in Elementary Geometry, is not general enough for the pur- poses of Trigonometry. We shall have to deal with positive and negative angles of any magnitude. Such a conception of angles may be formed as follows : An angle may be considered as generated by a line which first coin- cides with one side of the angle, then revolves about the vertex, and finally coincides with the other side. initial side This line is called the generating line of the angle. In its first position it is said to coincide with the initial side of the angle, and in its final position with the terminal side of the angle. Thus, the angle A OB is generated by the line OP revolving about in the direction indicated from the initial side OA to the terminal side OB. 10. Positive and negative angles. In the above figures the angles were generated by revolving the generating line counter-clockwise ; mathematicians have agreed to call such angles positive. Below are angles having the same initial and terminal sides as those above, but the angles are different since they have been generated by revolving the generating line clockwise ; such angles are said to be negative.* * The arcs with arrowheads will be drawn full when indicating a positive angle, and dotted when indicating a negative angle. 24 TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 25 11. Angles of any magnitude. Even if angles have the same initial and terminal sides, and have been generated by rotation in the same direction, they may be different. Thus, to generate one right angle, the generating line rotates into the position OB as shown in Fig. a. If, however, the generating line stops in the position OB after making one complete revolution, as shown in Fig. b, then we have generated an angle of magnitude five right angles ; or, if two com- plete revolutions were first made, as shown in Fig. c, then we have Cl Fie. a Fig. b Fig. c generated an angle of magnitude nine right angles ; and so on indefi- nitely. This also shows that positive angles may have any magnitude whatever. Similarly, by making complete revolutions clockwise, it is seen that negative angles may have any magnitude.* 12. The four- quadrants. It is customary to divide the plane about the vertex of an angle into four parts called quadrants, by passing two mutually perpendicular lines through the vertex. Thus, if is the vertex, the different quadrants are named as indicated in the figure below, the initial side being horizontal and drawn to the right. An angle is said to be (or lie) in a certain quadrant when its ter- minal side lies in that quadrant. In the figures shown on the pre- vious page, only the least positive and negative angles having the given initial and terminal sides are indicated by the arcs. As a matter of fact there are an infinite number of positive and negative angles in each case which have the same initial and terminal sides, all differing in magnitude by multiples of 360°. The following examples will illustrate the preceding discussion. Second Quadrant Third Quadrant First Quadrant initial side Fourth Quadrant > Thus, the minute hand of a clock generates - 4 rt. d every hour, i.e. - 96 rt. A every day. 26 PLANE TRIGONOMETRY EXAMPLES 1. Show that 1000° lies in the fourth quadrant. Solution. 1000° = 720° + 280° = 2 x 360° + 280°. Hence we make two com- plete revolutions in the positive direction and 280° beyond, and the terminal- side of 280° lies in the fourth quadrant. 2. Show that — 568° lies in the second quadrant. Solution. — 568° = — 360° — 208°. Hence we make one complete revolution in the negative direction and 208° beyond in the negative direction, and the terminal side of — 208° lies in the second quadrant. 3. In what quadrants are the following angles ? (a) 225°. (e) 651°. (i) 540°. (m) 1500°. (b) 120°. (f) - 150°. (j) 420°. (n) 810°. (c) - 315°. (g) - 75°. (k) - 910°. (o) - 540°. (d) -240°. (h) -1200°. (1) -300°. (p) 537°. 13. Rectangular coordinates of a point in a plane. In order to define the functions of angles not acute, it is convenient to introduce the notion of coordinates. Let A"'A" be a horizontal line and Y'Y a line per- pendicular to it at the point O. Any point in the plane of these lines (as P) is determined by its distance _l> j an d direction from each of the perpen- diculars A*'A' and Y'Y. Its distance from Y'Y (as NP = a) is called the abscissa of the point, and its distance from A" A (as MP = b) is called the ordinate of the point. Abscissas measured to the right of Y'Y are positive. Abscissas measured to the left of F'Fare negative. Ordinates measured above X'X are positive. Ordinates measured below X'X are negative. The abscissa and ordinate taken to- r ' gether are called the coordinates of the II I point and are denoted by the symbol (-»+) (+,+) (a, b). The lines X'X and Y'Y are called the axes of coordinates, X'X being the axis of abscissas or the axis ofX, and FT the axis of ordinates or the axis of Y; and the point O is called the origin of coordinates. The axes of coordinates divides the plane into four parts called quadrants (just as in the previous section), the figure indicating the proper signs of the coordinates in the different quadrants. x±±=i X' III X IV TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 27 \ ->X To plot a point is to locate it from its coordinates. The most con- venient way to do this is to first count off from along A" A' a num- ber of divisions equal to the abscissa, to the right or left according as the abscissa is positive or negative. Then from the point so determined -|— - count off a number of divisions equal ~ to the ordinate, upward or downward according as the ordinate is positive x ' or negative. The work of plotting points is much simplified by the use of coordinate or plotting paper, con- (V3) structed by ruling off the plane into equal squares, the sides being parallel to the axes. Thus, to plot the point (4, — 3), count off four divisions from on the axis of X to the right, and then three divisions downward from the point so determined on a line parallel to the axis of Y. Similarly, the following figures show the plotted points (- 2, 3), (- 3, - 4), (0, 3). tr 1 — Y' 1 > J =t JL- — 1 1— — t\ 1 Xf (-3(-41 T.. ■^II 1 -4- (6,4). *X 14. Distance of a point from the origin. Represent the abscissa of a point P by a and the ordinate by b, and its distance from the Yjl ' i i origin by h. Then h = Va 2 + b% since h is the hypotenuse of a right triangle whose sides are a and b. Although h may be either positive or negative, it will be sufficient for our purposes to treat it as being always positive. In order to become familiar with the notion of coordinates, the student should plot a large number of points. ■*-x 28 PLANE TRIGONOMETRY EXAMPLES 1. (a) Plot accurately the points (5, 4), (- 3, 4), (- 2, - 4), (6, - 1), (6, 0), (- 5, 0), (0, 4), (0, - 3). (b) What is the distance of each point from the origin ? Ans. Vil, 5, 2 VI, etc. 2. Plot accurately the points (1, 1), (- 1, - 1), (-1, 1), (Vz, 1), (Vz, - 1), ( — V5, — 1), and find the distance of each one from the origin. 3. Plot accurately the points (V2, 0), (- 6, - 10), (3, - 2 V5), (10, 3), (0, 0), (0, -V5), (3, -5), (-4, 5). 15. Trigonometric functions of any angle defined. So far the six trigonometric functions have been defined only for acute angles (§ 1, p. 2). Now, however, we shall give a new set of definitions which will apply to any angle whatever, and which agree with the definitions already given for acute angles. & Angle in first quadrant Y \ i >X &■ Yi < B B 1 3V P < » < 2 O ¥'■ X' +x Angle in second quadrant jf- n n l < \ , (, \ A / r* s A 3 1 ^\ B B r T ¥X Angle in third quadrant Angle in fourth quadrant Take the origin of coordinates at the vertex of the angle and the initial side as the axis of A'. Draw an angle XOB in each quadrant. From any point P on the terminal side OB of the angle draw PQ perpendicular to the initial side, or the initial side produced. In every case OQ is the abscissa and QP the ordinate of the point P. TBIGONOMETKIC FUNCTIONS OF ANY ANGLE 29 Denoting by XOB any one of these angles, their functions are defined as the following ratios : op ordinate op hypotenuse (10) smXOB = ^- = r ; ) (13) cscXOg = — »= .. x — ; v ' OP hypotenuse v ' QP ordinate (11) cos XOB = ^ = hypotenuse _ qp _ ordinate _ ^ ' OQ abscissa ' OP hypotenuse (14) secXOB = ^=-^—. ; v ' OQ abscissa 00 abscissa * (15) cotJCOg = ^= .. „ •* x ' QP ordinate To the above six functions may be added the versed sine (written ver- sin) and coversed sine (written coversin), which are defined as follows : veisia XOB = 1 — cos XOB ; coversin XOB = 1 — sin. XOB. 16. Algebraic signs of the trigonometric functions. Bearing in mind the rule for the algebraic signs of the abscissas and ordinates of points given in § 13, p. 26, and remembering that the hypotenuse OP is always taken as positive (§ 14, p. 27), we have at once, from the defi- nitions of the trigonometric functions given in the last section, that : In I Quadrant, all the functions are positive. In II Quadrant, sin and esc are positive ; all the rest are negative. In III Quadrant, tan and cot are positive; all the rest are negative. In IV Quadrant, sec and cos are positive ; all the rest are negative. These results are also exhibited in the following Eulb fok Signs II J VI sin + CSC + all + tan + cos + iirs cot + sec + A y All functions not indicated in each quadrant are negative. This rule for signs is easily memorized if the student remembers that reciprocal functions of the same angle must necessarily have the same sign, i.e. s in anc l esc have the same sign, cos and sec have the same sign, tan and cot have the same sign. 17. Having given the value of a trigonometric function, to construct geometrically all the angles which satisfy the given value, and to find the values of the other five functions. Here we will make use of the * As in acute angles it is seen that the functions in one column are the reciprocals of the functions in the other. 30 PLANE TRIGONOMETRY notion of coordinates, assuming as before that each angle has its vertex at the origin, and its initial side coinciding -with the axis of X. It remains, then, only to fix the terminal side of each angle, or, what amounts to the same thing, to determine one point (not the origin) in the terminal side. When one function only is given, it will appear that two terminal sides satisfying the given condition may be constructed. Thus, if we have given tanx = f,'we may write tan x = *x 2 — 2 _ ordinate — 3 abscissa ' (12), p. 29 Fig. a and hence, taking tan x = § , one ter- minal side is determined by the origin and (3, 2), giving the angle XOB (in the first quadrant). 2 — - j is determined by the The other terminal side, taking tan x origin and (— 3, — 2), giving the angle XOB' (in the third quadrant) Hence all the angles x * which satisfy the condition tan x = § have the initial side OX and the terminal side OB, or, have the initial side OX and the terminal side OB'. Fig. a Fig.J Let us now determine the values of all the functions. From Fig. a, OP = y/~OQ 2 + QP 2 = V9 + 4 = Vl3 (always positive). Hence by § 15, p. 29, from Fig. a, 2 sin XOB = cos XOB = taxi XOB = Vl3' 3 Vl3' 2 3 ! esc XOB = sec XOB = Vl3 2 ! Vl3 3 '' cot XOB = - ■ 2i * It is evident that, corresponding to each figure, there are an infinite number of both positive and negative angles differing by multiples of 360° which satisfy the given condition. TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 31 Similarly, irom Fig. b, sin XOB' ■■ 2 Vis' cos XOB' = - Vl3 tan X OB' = 3' esc XOB' = ■ sec XOB' = ■ cot XOB' = : Vl3 2 ! Vl3 Or, denoting by a; any angle which satisfies the given condition, we may write down these results in more compact form as follows : sin x = ± Vis' cos x = ± . — ; Vl3 tan x = — ; o csc x = ± sec a; = ± 3 cot a; = — ■ Vl3 2 5 Vl3 The method is further illustrated in the following examples : Ex. 1. Having given sin x = — \, construct the angle x. Also find the values of the other five functions. Solution. Here we may write,.by (10), p. 29, 1 _ — 1 _ ordinate 3 3 hypotenuse sin a; : (hypotenuse always positive). Since abscissa = ± V(hypot.) 2 — (ord.) 2 _= ± V9 — 1 = ± 2 V2, one terminal side is determined by the origin and ( — 2 V2, — 1) , ( y, , giving the angle XOB in the third quadrant. Here *X sin XOB = - cos XOB tan XO.B The other terminal side is determined by the origin and (2 V2, — 1) giving Y4- V csc XOB = - 3; 2V2 - ~3~ ; sec XOB = 2 V2 1 2 Vi' cot XOB = 2 V2. the angle XOB' in fourth quadrant. Here 1 *X sin XOR = - cos XOB' = tan XOB' = 3' 2j/2 ~3~ ; 1 2V2' cscXOB / = -3; 3 sec XOB' = ; 2V2 cotXaB' = -2\/2. 32 PLANE TRIGONOMETRY Or, denoting by x any angle which satisfies the given condition, we have sin x = ; esc x = — 6 ; 3 2V2 ,. 3 cos x = T —^— ; sec x = =F — -= , 3 2V2 tanx = ± ; cotx = ±2\/2. 2V2 771 Ex. 2. Having given cot x = — , find all the other functions of x. n Solution. Here we may write, by (15), p. 29, m — m abscissa cotx = — = = — , n — n ordinate and hypotenuse = Vm s + n 2 . Hence one terminal side is determined by the origin and (ra, n), and the other terminal side by the origin and (— m, — n). Therefore n Vm ! + n 2 sin x = ± ■ ; esc x = ± ; Vm" + n 2 n m Vm 2 + n 2 cos x = + — ; secx = ± i Vm 2 + n 2 ra tanx = — ; cotx = — • m » EXAMPLES In each of the following examples construct geometrically the angle x, and compute the values of all the functions of x. Given. 1. sin x = - • 5 1 2. cos x = 3 3. cotx = -3.t 5 4. sec x = 3 „ 13, 5. esc x = — 5 c * When m and n have the same sign, x represents angles in the first and third quadrants. "When m and n have opposite signs, x represents angles in the second and fourth quadrants. 3 3 t cot X = — 3 = = 1-1 6. a tan x = - ■ b 7. sin x = c. 8. a 2 -6 2 a 2 + 6 2 9. esc x ■= — V3. 10. m cos x = — • 11. tan x = — V 1 12. 2 sin x = 3 13. tan x = 2. 5. 14. sec x = p. TKIGONOMETRIC FUNCTIONS OF ANY ANGLE 33 ANSWERS Quad- rant sin cos tan CSC sec cot 1. I 3 5 4 5 3 4 5 3 5 4 4 3 II 3 5 4 5 3 4 5 3 5 4 4 ~3 2. II 2V^ 3 1 3 -2V2 3 2V2 -3 1 2 V2 HI 2V2 3 1 3 2V2 3 2 V2 -3 1 2V2 3. II 1 VTo 3 VTo 1 3 VTo VTo -3 IV 1 VTo 3 VTo 1 3 -VTo VTo 3 -3 4 3 4 5 5 3 4. II 5 5 ~3 4 3 i i 3 4 5 5 3 III 5 5 3 4 3 4 5 12 6 13 13 12 5. I 13 13 12 5 12 ~5~ 5 12 5 13 13 12 II 13 a _ 13 6 X 12 a 6 c 4. .. .. 5 12 If 6 Va 2 + 6 2 Va 2 +& 2 b Va 2 + 6 2 c Va 2 +6 2 ± Vl-c 2 a 1 c 1 6 1 Vl-c 2 a 7 Vl-c 2 Vl-c 2 c 8. I, 2 aft a 2 +6 2 a 2 -6 2 2 aft a 2 -ft 2 a 2 + 6 2 ± lab a 2 + 6 2 a 2 -& 2 ± 2a& IV a 2 + 6 2 a 2 -ft 2 9. III, IV 1 Vs *4 m c 1 -Vs c -1- c m ±V2 10 y/&-Tr& Vc 2 -m 2 1 m c m Vt?-m* Vc*-m 2 11. rv VTi 4 V2 -V7 4 VTi 4 1 V7 12. in, IV 2 3 V5 2 ±Vl 3 2 3 ¥ vi ^ 13. hi 5 V29 2 V29 5 2 V29 V29 ±_ 2~ 2 5 Vp 2 -1 1 p + * P 1 -1 14. ±Vp 2 -l Vp 2 -! Vp 2 -i 34 PLANE TEIGONOMETEY 18. Five of the trigonometric functions expressed in terms of the sixth. For this purpose it is again convenient to use the definitions of the functions which depend on the notion of coordinates ( § 13, p. 26). The following examples will illustrate the method. Ex. 1. Express, in terms of sin a;, the other five functions of x. sin x ordinate Solution. Since sin a; = • hypotenuse by (10), p. 29 abscissa = ± V (hypotenuse) 2 — (ordinate) 2 — ± Vl — sin 2 x. Hence, by definitions,* ±Vl-si» ! a: sin x = sin x ; sin a; cos x = ± Vl — sin 2 a; j tan x = ± - sec x = ± ■ smx vl — sin 2 x cot x = ± Vl — sin 2 x Vl — sin 2 x Ex. 2. Express, in terms of tan x, the other five functions of x. tan x _ ordinate 1 abscissa Solution. Since tan x : hypotenuse = ± V (abscissa) 2 + (ordinate) 2 Hence = ±Vl + tan 2 x. Vl+ tan 2 x esc x = ± ■ tanx Vl + tan 2 x tan x = tan x ; secx = ± Vl + tan 2 x; 1 cotx = tanx Ex. 3. Having given secx = f , find the values of the other five functions. „,..„. 5 hypotenuse , „ , x Solution. Since secx = - = ^^ , by (14), p. 29 4 abscissa ordinate = ± V(hypotenuse) 2 — (abscissa) 2 = ±V25-16 = ±3. Hence sin x = ± f ; cos x = J ; tan x = ± | ; cscx =± -j sec x = j ; cot x = ± $ • * It is convenient to draw a right triangle (as above) to serve as a check on the numer- ical part (not the algebraic signs) of our work. We then refer to the definitions of the functions of an acute angle (p. 2) where adjacent side corresponds to the abscissa, and opposite side corresponds to the ordinate. TK1G0N0METK1C FUNCTIONS OP ANY ANGLE 35 EXAMPLES 1. Express, in terms of cos x, the other five functions of x. Ans. sin x = ± Vl — cos 2 x ; cscx = ± cosx=cosx; secx = Vl — cos 2 x 1 cosx' Vl— cos 2 x cosx tanx=± ; cotx = ±- Vl — cos 2 x 2. Express, in terms of cotx, the other fire functions of x. Ans. sin x = ± — ; esc x = ± Vl + cot 2 x ; Vl+cot^x cotx Vl + cot 2 x cos x = ± — ; sec x = ± ; Vl + cot 2 x cotx tan x = ; cotx = cotx. cotx 3. Express, in terms of secx, the other five functions of x. Vsec 2 x — 1 sec x Ans. sin x = ± ; cscx = ± ■ secx Vsec 2 x-1 1 cos x = ; sec x = sec x : secx tanx = ± Vsec 2 x — 1 ; cotx = ± — • Vsec 2 x — 1 4. Express, in terms of esc x, the other five functions of x. Ans. sin x = ; esc x = esc x ; cscx Vcse 2 x — 1 cscx cosx = ± ; secx = ± oscx Vcse 2 x-1 tan x = 4- ; cotx = ± Vcsc 2 x — 1 . Vcsc 2 x — 1 5. Having given sec x = — ^-, find the values of the other five functions of x. Ans. sin x = ± |f ; cscx = ± |J; cos x = — ^y ; sec x = — lg- ; tanx = T "V"' cotx ~ ^ A- 6. Having given sin x = a, find the values of the other functions of x. Ans. sin x = a ; esc x = - ; a cos x = ± Vl - a? ; see x = ± — — Vl- a 2 a Vl-a 2 tan x = ± , ; cot x = ± ■ VT^Hi? a 36 PLANE TRIGONOMETRY 7. Having given cot x = V2, find the values of the other functions of x. Ans. sina; = ± — ; cscx = ±v3; cosx = ±^-; secs = ±-y-; tan x = ; cot x = V2. 19. Line definitions of the trigonometric functions. The definitions of the trigonometric functions given in § 15, p. 29, are called the ratio definitions. From these we shall now show how the functions of any C a- Angle in first quadrant Angle in second quadrant C -c Angle in third quadrant Angle in fourth quadrant angle may be represented by the numerical measures of the lengths of lines drawn as shown above in connection with a unit circle (i.e. a circle with radius unity). TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 37 Applying these ratio definitions, we get QP smAOP = 7Jp-^T ) = QP ' 00 cos AOP = -P-— = 00 : OP(=l) v ' txnAOP = ^ = AT * = AT; sec .lOP = OQ OA (= 1) OP_ OT * OQ ~ OA (= 1) cot .40P = -^ = — — — — ' = BC: QP OB (= 1) .„„ OP OC t esc .40P = = -— — ' = OC. QP OB (= 1) From these results the so-called line definitions of the trigono- metric functions may be stated as follows : The sin equals the length of the perpendicular drawn from the extremity of the terminal radius to the horizontal diameter. The cos equals the length of the line drawn from, the center to the foot of this perpendicular. The tan equals the length of a line drawn tangent to the circle from, the right-hand extremity of the horizontal diameter and meet- ing the terminal radius produced. The sec equals the distance from the center to the point of inter- section of this tangent with the terminal radius produced. The cot equals the length of a line drawn tangent to the circle from, the upper extremity of the vertical diameter and meeting the terminal radius produced. The esc equals the distance from the center to the point of intersec- tion of this cotangent with the terminal radius produced. Algebraic signs must, however, be attached to these lengths so as to agree with the rule for the signs of the trigonometric functions on p. 29. We observe that sin and tan are positive if measured upward from the horizontal diameter, and negative if measured downward ; cos and cot are positive if measured to the right of the vertical diam- eter, and negative if measured to the left ; sec and esc are positive if measured in the same direction as the terminal side of the angle, and negative if measured in the opposite direction. * Since triangles OQP and OAT are similar, t Since triangles OQP and OBC are similar. 38 PLANE TRIGONOMETBY 20. Changes in the values of the functions as the angle varies. (a) The sine. Let x denote the variable angle A OP. As x decreases, the sine decreases through the values QiP lt Q2P1, etc., and as x approaches zero as a limit, the sine approaches zero as a limit. This is written .,, . sin 0=0. As x increases from 0° and approaches 90° as a limit, the sine is positive, and increases from zero through the values Q S P S , QJPi, etc., and approaches OB (= 1) as a limit. This is written sin 90°= 1. As x increases from 90° and approaches 180° as a limit, the sine is positive and decreases from OB (= 1) through Q 5 P 5 , etc., and ap- proaches zero as a limit. This is written sin 180° = 0. As x increases from 180° and approaches 270° as a limit, the sine is negative and increases in nu- merical value from zero through Q s P e , etc., and approaches the limit OB' {=—!). This is written sin 270° = - 1. As x increases from 270° and approaches 360° as a limit, the sine is negative and de- creases in numerical value from OB' (= — 1) through Q 7 P 7 , etc., and approaches the limit zero. This is written sin 360° = 0. (b) The cosine. Using the last figure, we see that as x decreases, the cosine increases through the values OQ u OQ 2 , etc., and as x ap- proaches zero as a limit, the cosine approaches the limit OA (= 1). This is written „„ n o , cos 0=1. As x increases from 0° and approaches 90° as a limit, the cosine is positive and decreases from OA (= 1) through the values OQ 8 , OQ t) etc., and approaches the limit zero. This is written cos 90° = 0. TEIGONOMETKIC FUNCTIONS OF ANY ANGLE 39 As x increases from 90° and approaches 180° as a limit, the cosine is negative and increases in numerical value from zero through OQ 6 , etc., and approaches the limit OA'(=— 1). This is written cos 180° = — 1. As x increases from 180° and approaches 270° as a limit, the cosine is negative and decreases in numerical value from OA' (= — 1) through OQ 6 , etc., and approaches the limit zero. This is written cos 270° = 0. As x increases from 270° and approaches 360° as a limit, the cosine is positive and increases from zero through OQ 1} etc., and approaches the limit OA (= 1). This is written cos 360° = 1. (c) The tangent. Let x denote the variable angle AOT. As x decreases, the tangent decreases through the values A T 1} A T 2 , etc., and as x approaches zero as a limit, the tangent approaches the limit zero.' This is written tan 0° = 0. t As x increases from 0° and approaches 90° as a limit, the tangent is positive and increases from zero through the values AT % , AT 4 , etc., without limit, i.e. beyond any numerical value. This is written tan 90° = + co.* Now suppose the angle x to be equal . to the angle A OP and let it approach 90° as a limit ; then the corresponding tangent A T e is negative and increases in numerical value without limit. This is written tan 90° = — ao. We see, then, that the limit of the tangent will be + oo or — oo according as x is increasing or decreasing as it approaches the limit 90°- As one statement these last two results are written tan 90° = oo, when, as in this book, no distinction is made for the manner in which the angle approaches the limit 90°. * + oois read plus infinity. — oo is read minus infinity. oo is read simply infinity. 40 PLANE TRIGONOMETRY As x increases from 90° and approaches 180° as a limit, the tangent is negative and decreases in numerical value from — oo through AT 6 , AT h , etc., and approaches the limit zero. This is written tan 180° = 0. As x increases from 180° and approaches 270° as a limit, the tangent is positive and increases from zero through A T„, A T 4 , etc., without limit. This is written tan 270° = oo. As x increases from 270° and approaches 360° as a limit, the tangent is negative and decreases in numerical value from — oo through AT 6 , AT B , etc., and approaches the limit zero. This is written tan 360° = 0. (d) The secant. Using the last figure, we see that as x, decreases, the secant decreases through the values OT x , OT it etc., and approaches OA (= 1) as a limit. This is written sec 0° = 1. As x increases from 0° and approaches 90° as a limit, the secant is positive and increases from OA (= 1) through OT it OT i} etc., with- out limit. This is written sec 90° = oo. As x increases from 90° and approaches 180° as a limit, the secant is negative and decreases in numerical value from — oo through 01\, OT 5 , etc., and approaches minus OA (= — 1) as a limit. This is written sec 180° = — 1. As x increases from 180° and approaches 270° as a limit, the secant is negative and increases in numerical value from minus OA (= — 1) through OT s , OT i} etc., without limit. This is written sec 270° = oo. As x increases from 270° and approaches 360° as a limit, the secant is positive and decreases from + oo through OT 6 , OT 5 , etc., and ap proaches the limit OA (= 1). This is written sec 360° = 1. TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 41 (e) The cotangent. Let x denote the -variable angle AOC. As x decreases, the cotangent increases through the values BC lt BC 2 , etc., and as x ap- proaches 0° as a limit, the cotangent increases with- out limit. This is written cot0° QO. As x increases from 0° and approaches 90° as a limit, the cotangent is posi- tive and decreases from + co through the values BC S , BC it etc., and approaches the limit zero. This is written cot 90° = 0. As x increases from 90° and approaches 180° as a limit, the cotan- gent is negative and increases in numerical value from zero through BC S , BC S , etc., without limit. This is written cot 180° = oo. As x increases from 180° and approaches 270° as a limit, the cotan- gent is positive and decreases from + co through BC Z , BC it etc., and approaches the limit zero. This is written cot 270° = 0. As x increases from 270° and approaches 360° as a limit, the cotan- gent is negative and increases in numerical value from zero through BC h , BC e , etc., without limit. This is written cot 360° = oo. (/) The cosecant. Using the last figure, we see that as x decreases, the cosecant increases through the values OC^, OC 2 , etc., and as x approaches 0° as a limit, the cosecant increases without limit. This is written csc0° : OO. As x increases from 0° and approaches 90° as a limit, the cosecant is positive and decreases from + co through 0C S) 0C 4 , etc., and ap- proaches the limit OB(=l). This is written esc 90° = 1. 42 PLANE TKIGONOMETRY As x increases from 90° and approaches 180° as a limit, the cose- cant is positive and increases from OB (= 1) through OC b) OC & , etc., without limit. This is written esc 180° = oo. As x increases from 180° and approaches 270° as a limit, the cose- cant is negative and decreases in numerical value from — oo through 0C 8 , OCi, etc., and approaches the limit minus Ofi(=— 1). This is written esc 2 70° — — 1 . As x increases from 270° and approaches 360° as a limit, the cosecant is negative and increases in numerical value from minus OB(= — 1) through OC s , OC 6 , etc., without limit. This is written esc 360° = oo. These results may be written in tabulated form as follows : * 0° 90° 180° 270° 360° sin 1 -1 cos 1 -1 1 tan 00 00 cot oo oo oo sec 2 00 -1 oo 1 CSC oo 1 oo -1 CO It is of importance to note that as an angle varies its sine and cosine can only take on values between — 1 and + 1 inclusive ; tangent and cotangent can take on any values lohatever ; secant and cosecant can take on any values whatever, except those lying between — 1 and + 1. Prove the following : (a) sinO° + cos 90° = 0. (b) sin 180° + cos 270° = 0. (c) cosO° + tanO° = l. (d) tanl80° + cot90° = 0. (e) sin 270° - sin 90° = - 2. EXAMPLES (f) cos 0° + sin 90° = 2. (g) cos 180° + sin 270° = - 2. (h) sec0° + csc90° = 2. (i) sec 180° - sec 0° = - 2. (]') cos 90° -cos 270° = 0. (k) sin 90° + cos 90° + esc 90° + cot 90° = 2. (1) cos 180° + sec 180° + sin 180° + tan 180° = - 2. (m) tan 360° - sin 270° - esc 270° + cos 360° = 3. * The above table is easily memorized if the student will notice that the first four columns are composed of squares of four blocks each, in which the numbers on the diagonals are the same ; also the first two columns are identical with the next two if 1 be replaced by — 1 ; also the first and last columns arc identical. TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 43 2. Compute the values of the following expressions : (a) a sin 0° + 6 cos 90° - c tan 180°. Ans. 0. (b) a cos 90° -6 tan 180° + c cot 90°. 0. (c) a sin 90° - b cos 360° + (a - b) cos 180°. 0. (d) (a 2 - & 2 ) cos 360° - 4 ab sin 270°. a 2 + 4 ab - & 2 . 21. Angular measure. There are two systems in general use for the measurement of angles. For elementary work in mathematics and for engineering purposes the system most employed is Degree measure, or the sexagesimal system. * The unit angle is one degree, being the angle subtended at the center of a circle by an arc whose length equals ^^ of the circumference of the circle. The degree is subdivided into 60 minutes, and the minute into 60 seconds. Degrees, minutes, and seconds are denoted by symbols. Thus 63 degrees 15 minutes 36 seconds is written 63° 15' 36"- Reducing the seconds to the decimal part of a minute, the angle may be written 63° 15.6'. Reducing the minutes to the decimal part of a degree, the angle may also be written 63.26°. f It has been assumed that the student is already familiar with this system of measuring angles, and the only reason for referring to it here is to compare it with the following newer system. 22. Circular measure. The unit angle is one radian, being the angle subtended at the center of a circle by an arc whose length equals the length of the radius of the circle. Thus, in the figure, if the length of the arc AB equals the radius of the circle, then angle A OB = 1 radian. The circular measure of an angle is its magnitude expressed in terms of radians. This system was introduced early in the last century. It is now used to a cer- tain extent in practical work, and is universally used in the higher branches of mathematics. Both of the above systems will be used in what follows in this book. % * Invented by the early Babylonians, -whose tables of weights and measures were based on a scale of 60. This was probably due to the fact that they reckoned the year at 360 days. This led to the division of the circumference of a circle into 360 degrees. A radius laid off as a chord would then cut off 60 degrees. t To. reduce seconds to the decimal part of a minute we divide the number of seconds by 60. Similarly, we reduce minutes to the decimal part of a degree. See Conversion Tables on p. 17 of Granville's Four-Place Tables of Logarithms. X A third system is the Centesimal or French System. The unit is one grade, being ,J 5 of a right angle. Each grade is divided into 100 minutes and each minute into 100 seconds. This system has not come into general use. 44 PLANE TRIGONOMETRY Now let us find the relation between the old and new units. From Geometry we know that the circumference of a circle equals 2 ttR ; and this means that the radius may be measured off on the circumfer- ence 2 7r times.* But by the above definition each radius measured off on the circumference subtends an angle of one radian at the center, and we also know that the angles about equal 360°. Therefore 2 it radians = 360°, 7r radians = 180°, . ,. 180° 180° 1 radian = = , or, 7r 3.1416 (16) 1 radian = 57.2957° +. It therefore follows at once that : To reduce radians to degrees, multiply the number of radians by 57.2957 {=f} To reduce degrees to radians, divide the number of degrees by 57.2957 (=^)- Since 360 degrees = 2 tt radians, it ,. 3.1416 ,. 1 degree = — - radian = radian, or, loO lou (17) 1 degree = .01745 radian. Hence the above rules may also be stated as follows : To reduce radians to degrees, divide the number of radians by ■ 01746 ( = iio)- To reduce degrees to radians, multiply the number of degrees by • 01745 ( = iio)- The student should now become accustomed to expressing angles in circular measure, thus : TT 360° = 2 7r radians, 60° = — radians, o 180° = ir radians, 30° = — radians, 6 ' 90° = — radians, 45° = ^ radians, 3tt IT 270° = — radians, 15° = — radians, etc. * The student should carefully observe that we do not lay off these radii as chords. TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 45 When writing the trigonometric functions of angles expressed in circular measure it is customary to omit the word " radians," thus : sin (it radians) is written simply sin tt and = sin 180°, /7T \ 7T tan / — radians I is written simply tan — and = tan 90°, cot/-— radians j is written simply cot — and = cot 135°, 4 '•< I -r- radians J is written simply cos — and = cos 150°, esc (1 radian) is written simply esc 1 and = esc 57.29°, sec (J- radian) is written simply sec ^ and = sec 28.65°, etc. Since the number of times that the radius of a circle can be meas- ured off on an arc of the same circle determines the number of radians in the angle subtended at the center by that arc, we have ,„„. ,_ , , ,. . , length of subtending arc (18) Number of radians in angle = — = ; — rr- 2 v / length of radius Hence, knowing any two of the three quantities involved, the third may easily be found. Ex. 1. What is the circular measure of the angle subtended by an arc of length 3. 7 in. if the radius of the circle is 2 in. ? Also express the angle in degrees. Solution. Substituting in (18), we have 3 7 Number of radians = -^— — 1.85. Ans. 2 To reduce this angle to degrees, we have, from (16), 1.85 x 57.2957° = 105.997°. Ans. Ex. 2. What is the radius of a circle in which an arc of length 64 in. subtends an angle of 2.5 radians? 64 Solution. Substituting in (18), 2.5 = — , R E = 25.6 in. Ans. EXAMPLES 1. In what quadrant does an angle lie * if its sine and cosine are both nega- tive ? if sine is positive and cosine negative ? if sine is negative and cosine posi- tive ? if cosine and tangent are both negative ? if cosine is positive and tangent negative ? if sine and cotangent are both negative ? if sine is negative and secant positive ? 2. What signs must the functions of the acute angles of a right triangle have? Why? * That is, in what quadrant will its terminal side lie 1 46 PLANE TRIGONOMETRY 3. What functions of an angle of an oblique triangle may be negative ? Why f 4. In what quadrant do each of the following angles lie ? 6tt it lie Utt l\ir 15 tt it + 2 3tt + 2 1 _ _6 ."l2 ; _ ; _ T ; ^" ; 4 ; 16 ; 6 ; 5- ' ' 1' '2' 5. Determine the signs of the six trigonometrical functions for each one of the angles in the last example. 6. Express the following angles in degrees : ,,.1.2'. 25 . 8t.«- + 1. 3 . 28. 3T + 2 Ans. 74.4844°; 28.6478°; 120°; -143.239°; -67.5°; 39.549°; -171.887°; -160.4279°; 130.92°. 7. Express the following angles in circular measure: 22 J°; 60°; 135°; -720°; 990°; -120°; -100.28°; 45.6°; 142° 43.2'; -243.87°; 125° 23' 19" (1°=. 01 745333). Ans. 0.3926; 1.0470; 2.3558; -12.5640; 17.2755; -2.0940; -1.7499; .7957; 2.4905; -4.2555; 2.1880. 8. Express in degrees and in radians : (a) Seven tenths of four right angles. (b) Five fourths of two right angles. (c) Two thirds of one right angle. Ans. (a) 252°, ?JL; (b)225°,^; (c) 60°, £. 5 4 3 9. Eind the number of radians in an angle at the center of a circle of radius 25 ft., which intercepts an arc of 37^ ft. Ans. 1.5. 10. Find the length of the arc subtending an angle of 4£ radians at the center of a circle whose radius is 25 ft. Ans. 112^ ft. 11. Find the length of the radius of a circle at whose center an angle of 1.2 radians is subtended by an arc whose length is 9.6 ft. .4ns. 8 ft. 12. Find the length of an arc of 80° on a circle of 4 ft. radius. Ans. 5.6 ft. 13. Find the number of degrees in an angle at the center of a circle of radius 10 ft. which intercepts an arc of 5 it ft. Ans. 90°. 14. Find the number of radians in an angle at the center of a circle of radius 3-t 2 t inches, which intercepts an arc of 2 ft. Ans. 7.64. 15. How long does it take the minute hand of a clock to turn through — if radians ? 50 Ans. — mm. IT 16. What angle in circular measure does the hour hand of a clock describe in39min.22isec? A - II rad. 64 17. A wheel makes 10 revolutions per second. How long does it take to turn through 2 radians, taking tt = V ? Ans. ^J^ sec. 18. A railway train is traveling on a curve of half a mile radius at the rate of 20 mi. per hour. Through what angle has it turned in 10 sec. ? Ans. 6-j*r degrees. 19. The angle subtended by the sun at the eye of an observer is about half a degree. Find approximately the diameter of the sun if its distance from the observer be 90,000,000 mi. Ans. 785,400 mi. TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 47 23. Reduction of trigonometric functions to functions of acute -angles. The values of the functions of different angles are given in trig- onometric tables, such, for instance, as the one on p. 9. These tables, however, give the trigonometric functions of angles between 0° and 90° only, while in practice we sometimes have to deal with positive angles greater than 90° ' and with negative angles. We shall now show that the trigonometric functions of an angle of any magnitude whatever, positive or negative, can be expressed in terms of the trigonometric functions of a positive angle less than 90°, that is, of an acute angle. In fact, we shall show, although this is of less importance, that the functions of any angle can be found in terms of the functions of a positive angle less than 45°. In the next eighteen sections x and y denote acute angles. 24. Functions of complementary angles. To make our discussion complete we repeat the following from p. 3. Theorem. A function of an acute angle is equal to the co-function of its complementary acute angle. , Ex. Express sin 72° as the function of a positive angle less than 45°. Solution. Since 90° — 72° = 18°, 72° and 18° are complementary, and we get sin 72° = cos 18°. Ans. EXAMPLES 1. Express the following as functions of the complementary angle : (a) cos 68°. (e) cot 9. 167°. (i) esc 62° 18'. (b) tan 48.6°. (f) sin 72° 61' 43". .. oot .2_7r (c) sec 81° 16'. tr U ' ' 6 ' (g) C °V (k) sin 1.2. K) Sm 3 ' (h) sec 19° 29.8'. (1) tan 66° 22.3'. 2. Show that in a right triangle any function of one of the acute angles equals the co-function of the other acute angle. 3. If A, B, C are the angles of any triangle, prove that sin£A = cos %(B + C). 25. Reduction of functions of angles in the second quadrant. First method. In the unit circle whose center is O (see figure on next page), let A OP' be any angle in the second quadrant. The func- tions of any such angle are the same as the corresponding functions of the positive angle AOP' = 180° - P'OQ'. Let x be the measure of the acute angle P'OQ', and construct AOP = P'OQ' = x. 48 PLANE TKIGONOMETKY Now draw the lines representing all the functions of the supple- mental angles x and 180° — x. From the figure angle QOP = angle P'OQ', OP = OP'. by construction equal radii Therefore the right triangles OPQ and OP'Q' are equal, giving OQ'=OQ. But OQ'= cos (180° - x) and OQ = cos x ; hence cos (180° — x) equals cos x in numerical value. Since they have opposite signs, however, we get cos (180° — x) = — cos x. Also, from the same triangles, Q'p' = QP. But Q'P' = sin (180° — x) and QP = sin x, and since they have the < B ^\ P/ T /r A 1 qr 2 V' same sign, we get sin (180° — x) = sin x. Similarly, the two right triangles OTA and OT'A may be proven equal, giving or, AT' = AT and OV = OT, tan (180° — a;) = — tan x and sec (180° — x) = — sec x. In the same manner, by proving the right triangles OBC and OBC' equal, we get or, BC = BC and OC = OC, cot (180° — x) = — cot a; and esc (180° — x) = esc x. esc (180° — x) = esc a; ; sec (180° — a;) = — sec x ; cot (180° — x) = — cot a;. Collecting these results, we have sin (180° — x) = sin x ; cos (180° — x) = — cos a; ; tan (180° — x) = — tan a; ; Hence we have the Theorem. The functions of an angle in the second quadrant equal numerically the same-named functions of the acute angle between its terminal side and the termdnal side of 180°. The algebraic signs, however, are those for an angle in the second quadrant. TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 49 Ex. 1. Express sin 123° as the function of an acute angle, and find its value. Solution. Since 180° - 123° = 57°, sin 123° = sin (180° - 57°) = sin 57° = .8387 (p. 9). Ans. Ex. 2.' Find the value of sec 6 Solution, sec — = sec 150° = sec (180° — 30°) = - sec 30° = Ans. 6 K ' V3 Ex. 3. Find tan 516°. Solution. 516° is an angle in the second quadrant, for 516° — 360° = 156°. Hence tan 516° = tan 156°* = tan (180° - 24°) = - tan 24° = - .4452. Ans. Second method. The angle A OP' may also be written 90° -f y, where y measures the acute angle BOP'. Since the angles BOP' and P'OQ' are com- plementary, we have, from theorem on p. 47, sin x = cos y ; esc x = sec y ; cos x = sin y ; secx=cscy; tanx=coty; cotx = tany. Since 180° — x = 90° + y, we get, combining the above results with the results on the previous page, sin (90° + y) = cos y ; esc (90° + y) = sec y ; cos (90° + y) = — sin y ; sec (90° + y) = — esc y ; tan (90° + V) = - cot y ; cot (90° + y) = — tan y. Hence we have the Theorem. The functions of an angle in the second quadrant equal numerically the co-named functions of the acute angle between its terminal side and the termi- nal side of 9<P. The algebraic signs, however, are those for an angle in the second quadrant. Ex. 4. Find the value of cos 109°. Solution. Since 109° = 90° + 19°, cos 109° = cos (90° + 19°) = - sin 19* = - .3256. Ans. Ex. 5. Find the value of cos — — ■ 4 Solution. — = 855° = 720°-+ 135°. 4 Therefore cos — = cos 855° = cos 135° = cos (90° + 45°) = - sin45° = • Ans. 4 V2 The above two methods teach us how to do the same thing, namely, how to find the functions of an angle in the second quadrant in terms of the functions of an acute angle. The first method is generally to be preferred, however, as the name of the function does not change, and hence we are less likely to make a mistake. * The above theorem was proven for an angle of any magnitude whatever whose termi- nal side lies in the second quadrant. The generating line of the angle may have made one or more complete revolutions before assuming the position of the terminal side. In that case we should first (if the revolutions have been counter-clockwise, i.e. in the positive direction) subtract such a multiple of 360" from the angle that the remainder will be a positive angle less than 3C0°. 50 PLANE TRIGONOMETRY EXAMPLES 1. Construct a table of sines, cosines, and tangents of all angles from 0° to 180° at intervals of 30°. Ans. 0° 30° 60° 90° 120° 150° 180° sin cos tan 1 1 2 V3 2 1 Vs V3 2 1 2 V3 1 CO V3 2 1 2 -V3 1 2 V3 2 1 "Vs -1 2. Construct a table of sines, cosines, and tangents of all angles from 90° to 180° at intervals of 15°, using table on p. 9. Ans. 90° 105° 120° 135° 150° 165° 180° sin cos tan 1.0000 0.0000 CO .9659 - .2588 - 3.7321 .8660 - .5000 -1.7321 .7071 -.7071 -1.0000 .5000 -.8660 -.5774 .2588 -.9659 -.2679 0.0000 -1.0000 0.0000 3. Construct a table of sines, cosines, and tangents of all angles from 90° to 135° at intervals of 5°. 4. Express the following as functions of an acute angle : (a) sin 138°. (b) tan 883°. 4 7T (e) cot v ' 5 (h) . 13ir sin 5 (c) cos 165° 20'. (f) cot 170.48°. « cos 2.58. (d) sec 102° 18'. (g) esc 317° U) tan 1.5. 5. Find values of the f ollowing : (a) sin 128° = .788. (b) cos 160°= -.9397. . . . Sir (g) sin—. (m) (n) cot 95° 14'. esc 126° 42.8' (c) tanl35°=-l. (d) sec— = -2. v ' 3 / \ . llir .. (e) cot = — 1. (h) tan 108° 15'. (i) cos 173° 9.4'. (j).tan \ti 6 (k) cos 496.7°. (o) (P) (q) (r) . 7tt sm 9 cos 500°. tan 870°. sec 1.9°. (f) esc 835°= 1.1034. (1) sec 168.42°. W tanl. 6. Express the following as functions of an acute angle less than 45° (a) sin 106° = cos 16°. llrr- (b) cos 148.3° = -cos 31.7°. ( e ) csc^-. (c) tan 862°. ^ 23? (d) sec 794° 52'. .„. -GO TV (f ) COS TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 51 sec x ; cot (180° + a) = cot a;. 26. Reduction of functions of angles in the third quadrant. First method. In the unit circle whose center is 0, let A OP' be any angle in the third quadrant. The functions of any such angle are the same as the corresponding functions of the positive angle A OP' = 180° + Q'OP'. Let * be the measure of the acute angle Q'OP', and construct AOP — Q'OP' = x. Now drawing the lines representing all the functions of the angles x and 180° + x, we get, just as in the previous case, sin (180° + x) = — sin x ; esc (180° + x) — — esc x ; cos (180° + x) — - cos x ; sec (180° + x) -. tan (180° + x) = tan x ; Hence we have the Theorem. The functions of an angle in the third quadrant equal numerically the same-named func- tions of the acute angle between its terminal side and the terminal side of 180°. The algebraic signs, how- ever, are those for an angle in the third quadrant. Ex. 1. Express cos 217° as the func- tion of an acute angle, and find its value. -B ' Solution. Since 217° - 180° = 37°, cos 217° = cos (180° + 37°) = - cos 37° = - .7986. Ans. Ex. 2. Find value of esc 225°. Solution, esc 225° = esc (180° + 45°) = -esc 45° = -V2. Ans. Ex. 3. Find value of sin 600°. Solution. 600° is an angle in the third quadrant, for 600° - 360° = 240°. V3 Hence sin 600° = sin 240° = sin (180° + 60°) = - sin 60° = - — . Ans. Second method. The angle A OP' may also be. written 270° - y, where y meas- ures the acute angle P'OB'. Since the angles P'OB' and Q/OP' (=AOP) are complementary, we have, from theorem on p. 47, combined with the above results, remembering that 180° + x = 270° - y, sin (270° - y) = - cos y ; esc (270° - y) = - sec y ; cos (270° - y) = - sin y ; sec (270° - y) = - esc y ; tan (270° - y) = cot y ; cot (270° - y) = tan y. Hence we have the Theorem. The functions of an angle in the third quadrant equal numerically the co-named functions of the acute angle between its terminal side and the termi- nal side of 270°- The algebraic signs, however, are those of an angle in the third quadrant. ^X P/ T 1 Q ' 1 rfx\ A V/6 Q / V* p R=l . 52 PLANE TRIGONOMETRY Ex. 4. Find sin 269°. Solution. Since 270° - 11° = 259°, sin 259° = sin (270° - 11°) = - cos 11° = - .9816. Ans. As in the last case, the first method is generally to be preferred. EXAMPLES 1. Construct a table of sines, cosines, and tangents of all angles from 0° to 270° at intervals of 45°. Ans. 0° 45° 90° 135° 180° 225° 270° sin 1 1 1 V2 1 -1 cos 1 l 1 "V2 -1 1 V5 tan l 00 -1 l oo 2. Construct a table of sines, cosines, and tangents of all angles from 180° to 270° at intervals of 15°, using table on p. 9. Ans. 180° 195° 210° 225° 240° 255° 270° sin cos tan - 1.0000 - .-2588 - .9659 .2679 - .5000 - .8660 .5774 -.7071 -.7071 1.0000 - .8660 -.5000 1.7321 - .9659 - .2588 3.7321 - 1.0000 00 3. Construct a table of sines, cosines, and tangents of all angles from 136° to 270° at intervals of 5°. 4. Express the following as functions of an acute angle : (a) tan 200°. (b) sin 583°. (c) cos 224° 26'. (d) sec 260° 40'. 5. Find values of the following : (a) tan 235° = 1.4281. (b) cot 1300° = 1.1918. (c) sin 212° 16'. ,,. 4ff 1 (d) cos — = V ' 3 2 . . 7tt 2 (e)sec~=- — . (j) csc (f) sin 609°. (e) cot o (f) csc 4.3. (g) sin 128°. 13tt (g) cos ^-- (h) tan 4. ... 29 7T (l) cot—-. 21 7T (k) sin 228.4°. (1) tan 255° 27. (h) cos 998.7°. ... . 16tt (i) sin—. ... 8tt (3) cos -g-- (m) cot 185° 52'. (n) cos 587°. (o) esc 7 (p) sin 262° 10'. (q) cos 204.86°. (r) tan TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 53 6. Express the following as functions of an acute angle less than 45° : , x 17tt (a) cos — . (b) tan 236.5°. (c) sin 594°. ,,, 5 JT (d) sec 23 7T (e) sin- i (f) cos 260° 53.4'. 27. Reduction of functions of angles in the fourth quadrant. First method. As before, let A OP 1 be any angle in the fourth quadrant. The functions of any- such angle are the same as the C^ corresponding functions of the positive angle A OP' = 360° — P'OQ. Let x be the measure of the acute angle P'OQ, and con- struct A OP = P'OQ = x. Now, drawing the lines representing all the functions of the angles x and 360° — x, we get, just as in the previous eases, sin (360° - as) = — sin x ; cos (360° — x) = cos x ; tan (360° - *) = — tan x ; esc (360° — x) = — esc x ; sec (360° — x) = see x ; cot (360° — x) = — cot x ; Hence we have the Theorem. The functions of an angle in the fourth quadrant equal numerically the same-named functions of the acute angle between its terminal side and the terminal side of 360°. The algebraic signs, however, are those for an angle in the fourth quadrant. Ex. 1. Express sin 327° as the function of an acute angle, and find its value. Solution. Since 360° - 327° = 33°, sin 327° = sin (360° - 33°) = - sin 33° = - . 5446. Ans. Ex. 2. Find value of cot — • o Solution, cot— = cot 300°= cot (360° -60°) = -cot 60° = - — • Ans. Ex. 3. Find value of cos 1000°. Solution. This is an angle in the fourth quadrant, for 1000° - 720° = 280°. Hence cos 1000° = cos 280° = cos (360° - 80°) = cos 80° =. 1736. Ans. 54 PLANE TRIGONOMETRY Second method. The angle A OP' may also be written 270° + y, where y measures the acute angle B'OP'. Since the angles B'OP' and P'OQ are com- plementary, we have, from theorem on p. 47, combined with the above results, remembering that 360° - x = 270° + y, sin (270° + y) =- cosy; cos (270° + y) = siny; tan (270° + y) =-coty; esc (270° + y) = - sec y ; sec(270° + ^) = oscy; cot(270° + y) = -tanjr. Hence we have the Theorem. The functions of an angle in the fourth quadrant equal numerically the co-named functions of the acute angle between its terminal side and the termi- nal side of 270°. The algebraic signs, however, are those of an angle in the fourth quadrant. IItt Ex. 4. Find value of cos ; Solution, cos— = cos 330°= cos (270° +60°) = sin 60° = — • Ans. 6 ^ As before, the first method is generally to be preferred. EXAMPLES 1. Construct a table of sines, cosines, and tangents of all angles from 180° to 360° at intervals of 30°. Ans. 180° 210° 240° 270° 300° 330° 360° 1 V3 Vs 1 sin ~2 2 — 1 2 2 V3 1 1 V3 1 cos — 1 2 2 2 2 tan 1 V3 V^ 00 -V3 1 ~V3 2. Construct a table of sines, cosines, and tangents of all angles from 270° to 360° at intervals of 15°, using table on p. 9. Ans. 270° 285° 300° 315° 330° 345° 360° sin cos tan -1.0000 00 -.9659 .2588 -3.7321 -.8660 .5000 -1.7321 -.7071 .7071 -1.0000 -.5000 .8660 -.5774 -.2588 .9659 -.2679 1.0000 3. Construct a table of sines, cosines, and tangents of all angles from 270° to 860° at intervals of 5°. , TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 55 4. Express the following as functions of an acute angle : (a) sin 289°. (b) cos 322.4°. (c) tan 295° 43'. (d) cot 356° 11'. • 5. Find values of the following : (a) sin275° = -.9962. (b) cos 336° = .9135. (c) tan 687° = -.6494. (d) cot 1055°. (e) sec 295° 52.6'. ... . 17 TT (f) sin v ; 9 (e) sm655°. /« ' 9 "" (f) esc — - (g) sin 275. 5°. (g) esc 5.2. (h) cos 4 ... 11 ir (i) csc-— - (h) cos 18 1 (j) fe& 5tt (i) sec 246°. 0) tan — - (k) sin 27 5° 22'. (1) cot 348°. (m) tan 660°. / x 13,r (n) sec v ' 8 . . . hir (o) sin — . 28. Reduction of functions of negative angles. Simple relations exist between the functions of the angle x and — x where x is any angle whatever. It is evident that x and — x will lie, one in the first quad- rant and the other in the fourth quadrant, as angles A OP and A OP' B „, , B in the right-hand figure ; or, one will lie in the second quadrant and the other in the third quadrant, as the angles A OP and A OP' in the left-hand figure. In either figure, remembering the rule for signs (§ 16, p. 29), we get QP=-QP', OQ = OQ, AT = -AT', OT = OT', BC=- BC, OC =-OC, . sin x = — sin (— x) ; cos x = cos (— x) ; , tan x = — tan (— x) ; . sec x = sec (— x) ; . cot x = — cot (— x) ; , csc x = — csc (— x). We may write these results in the form sin (— x) = — sin x ; esc (— x) — — csc x ; cos (— x) = cos x ; sec (— x) = sec x ; tan (— x) = — tan x ; cot(— x) = — cot x. 56 PLANE TRIGONOMETRY Hence we have the Theorem. The functions of — x equal numerically the same-named functions of x. The algebraic sign, however, will change for all functions except the cosine and secant* Ex. 1. Express tan ( — 29°) as the function of an acute angle, and find its value. Solution, tan (-29°) = -tan 29° = -.5543. Ans. Ex. 2. Find value of sec (- 135°). Solution, sec ( - 135°) = sec 135° = sec (180° - 45°) = - sec 46° = - Vi. Ans. Ex. 3. Find value of sin (- 540°). Solution. sm(-540°) = -sin54Q° = -sin(360 o +180 o ) = -sinl80°=0. Ans. EXAMPLES 1. Construct a table of sines, cosines, and tangents of all angles from 0° to 360° at intervals of 30°. Ans. 2. Find values of the following : (a) tan (-33°) = -.6494. V3 (b) sin(-60°) = -— • V2 (c) cos (-135°) = - (d) cot(- 259°). (e)sec(-j). (f) sin (-1231°). to) «*(-!£). (h) sin (-1000°). (i) cos (-2.3). M-t)- (k) sin (-176.9°). (1) cos (-88° 12.7') (m) tan/--^V (n) cot(- 842°). * Another method for reducing the functions of a negative angle consists in adding such a multiple of + 360° to the negative angle that the sum becomes a positive angle less than 360°. The functions of this positive angle -will be the same as the functions of the given negative angle, since their terminal sides will coincide. To illustrate : Ex. Find value of cos (- 240°). Solution. Adding + 360° to - 240° gives + 120°. Hence cos(- 240°)= cos 120° = cos (180° -60°)= -cos 60°=- J. Ans. Angle sin cos tan 0° 1 1 Vs 1 — 30° 2 2 "Vi -60° _Vs 2 1 2 -Vs -90° -1 00 -120° VI 2 1 _ 2 V5 - 150° 1 2 Vs 2 1 V§ - 180° -1 -210° 1 2 _V3 2 1 "vs -240° Vs 2 1 ~2 -Vs -270° 1 00 -300° Vs 2 1 2 V3 -330° 1 2 VI 2 1 Vs -360° 1 TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 57 29. General rule for reducing the functions of any angle to the functions of an acute angle. The results of the last seven sections may be stated in compact form as follows, x being an acute angle.* General Rule. I. Whenever the angle is 180° ±x or 360° ± x, the functions of the angle are numerically equal to the same-named functions of x. II. Whenever the angle is 90° ± x or 270° ± x, the functions of the angle are numerically equal to the co-named functions of x. III. In any case the sign of the result is the same as the sign of the given function taken in the quadrant where the given angle lies. The student is advised to use I wherever possible, since the liabil- ity of making a mistake is less when the name of the function remains unchanged throughout the operation. Work out examples from pp. 50-56, applying the above general rule. Angle sin cos tan cot sec CSC 90° 95° 100° 1.0000 .9962 .9848 0.0000 - .0872 -.1736 00 -11.430 -5.6713 0.0000 -.0875 -.1763 CO -11.474 -5.7688 1.0000 1.0038 1.0154 EXAMPLES 1. Construct a table for every five degrees from 90° to 180°. Ans. 2. Construct a table as in Ex. 1 for every 15° from 180° to 270°. 3. Construct a table as in Ex. 1 for every 10° from 0° to — 90°. 4. Reduce the following to functions of x : (a) sin (x— 90°) = — cosx.f (b) cos (x — 7r) = — cos x. (c) tan I — x 1 = cot x. (d) cot(x - 2tt) = cotx. (j) cot(- x - 8tt). (e) sec (x — 180°) = — sec x. (f ) esc ( — X ] = — sec x. (g) sin (x - 270°). (h) cos(— x — ir). (i) tan^x--^j- (k) sec(x-630°). (1) esc (x - 720°). * In case the given angle is greater than 360° we assume that it has first heen reduced to a positive angle less than 360° hy the subtraction of some multiple of 360°. Or, if the given angle is negative, "we assume that it has been reduced to a positive angle by the theorem on p. 66. t Since x is acute, % — 90° w a negative angle. Hence sin (x - 90°) = — sin (90° — x) = — cos a* 58 PLANE TRIGONOMETRY 5. Find values of the following : (a) cos 420°. (i) cot 549° 39'. , p)seo !£5. (b) sin7«8°. (j) esc 387° 68'. v/ 6 / 4ir\ „, / 11 t\ (q) sin (-5.3). (0)B ec(- T j. (k)sec(-— j. (d) cot (-240°) (r) cos ( — 1 • n tan /_ Stt\ W • \ 12/ (e)csc^. ^ i ' <s)tan(-^V (6) cso 3 (m) sin (-830°). W \ 12/ (f) tan 7. 5. 9ir (t) sec (- 123.8°). (g) sin (-2.8). (n) cos — . (u) sin (- 256° 19.6') (h) cos 952.8°. (o) cot 1020°. (v) cos(- 98° 3r). 6. Prove the following : (a) sin 420° • cos 390° + cos ( - 300°) • sin ( - 330°) = 1. (b) cos 570° • sin 510° - sin 330° • cos390° = 0. (c) a cos (90° -z) + b cos (90° + x) = (o - 6) sin x. (d) m cos I a;) ■ sinl x) = msinx cosx. (e) (a - 6) tan (90° - x) + (a + 6) cot (90° + x) = (a - b) cotx - (a + 6) tan x. (f) sinl — 1- x]sin(ir + x) + cosl — (- x)cos(7r — x) = 0. (g) cos(7T+x)cosl y\ — sm(7r + x)sml y\ = co&xsmy — smxcosy. (h) tan a; + tan(— y) — tan(-7r — y) = tanx. (i) cos (90° + a) cos (270° - a) - sin (180° - a) sin (360° - a) = 2 sin^a. sin (180° -y) 1 u; sin (270° -y) v "' sin2(270°-2/) (k) 3 tan 210° + 2 tan 120° = - V3. (n) tan ^ (2 w + x) = tan ^ x. (1) 5 sec" 135° - 6 cotf 300° = 8. (o) esc J (x - 2 n-j i = - sec \ x. (m) cos £ (x - 270°) = sin \ x, (p) cos ^ (y - 810°) = - sin £ y. CHAPTEE III RELATIONS BETWEEN THE TRIGONOMETRIC FUNCTIONS 30. Fundamental relations between the functions. From the defi- nitions (and footnote) on p. 29 we have at once the reciprocal relations sin* 1 (19) sin x = , esc x CSC* = (20) 1 cos X = , ^ sec* sec* = (21) 1 tan* = , rnt r cot* = COS* 1 tan*' Making use of the unit circle, we shall now derive five more very- important relations between the functions. In the right triangle QPO QP , , OQ tan x = — — i and cot x = — ■ Substituting the functions equivalent to QP and OQ, we get , s sin* cos* (22) tan-* = , cot* = - v ' cos* sin* Again, in the same triangle, B cot /^ N(5^ T / 44 a f f t&x\ COS A I Q v_ R=\ (23) In triangle OAT, (24) In triangle OCB, (25) QP +OQ =OP , or, sin 2 * + cos 2 * = 1. OA 2 + Zf 2 = Of 2 , or, 1 + tan 2 *=sec 2 *. OB* + BC 2 = OC\ or 1 + cot 2 * = csc 2 *. While in the above figure the angle a- has been taken in the first quadrant, the results hold true for any angle whatever, for the above 59 60 PLANE TRIGONOMETRY proofs apply to any one of the figures on p. 36 without the change of a single letter. While it is of the utmost importance to memorize formulas (19) to (25), p. 59, as they stand, the student should also learn the following formulas where each one of the functions is expressed explicitly in terms of other functions. (26) sin x = v ' cscr (19), p. 59 (27) sin x = ± Vl — cos 2 x* Solving (23), p. 59, for sin x (28) cos x = v ' sec x (20), p. 59 (29) cos x — ± Vl — sin 2 Jr. Solving (23), p. 59, for cos x (30) tan* = '• ' v ' cot X (21), p. 59 (31) tan x = ± Vsec 2 x — 1. Solving (24), p. 59, for tan x sin x sin x =fc Vl — cos 2 x (3°) tan.r — — — ' v ' cos at iVl-sin 2 * cos* [From (22), P- 59 ; also (29) and (27)-] (33) csc x = — v : ' sin x (19), p. 59 (34>. csc x = ± Vl + cot 2 x. Solving (25), p. 59, for csc x (35) sec x = v J cos x (20), p. 59 (36) sec x = ± Vl + tan 2 x. Solving (24), p. 59, for sec x 1 (37) cot x = v ' tan x (21), p. 59 (38) cot x = =fc Vcsc 2 x — 1. Solving (25), p. 59, for cot x , „ cos x cos x ± Vl — sin 2 x (39) cot X = = = ; . sinx ±Vl-cos 2 * Slnx [From (22), p. BS ; also (27) and (29).] 31. Any function expressed in terms of each of the other five functions. By means of the above formulas we may easily find any function in terms of each one of the other five functions as follows : * The double sign means that we get two values for some of the functions unless a con- dition is given which determines whether to choose the plus or minus sign. The reason for this is that there are two angles less than 360° for which a function has a given value. TRIGONOMETRIC RELATIONS 61 Ex. 1. Find sin x in terms of each of the other five functions of x. (a) SlnX = cic"£' from (26) (b) sin x = ± Vl - cos 2 a, from (27) (c) sini = substitute (34) in (a) ±V l + cot2a; (d) sinx = ± -Jl L_ = ± sec 2 x-l substitute (28) in (b) \ sec 2 a; sec a; (e) sin a: = = ta " X ■ Substitute (37) in (c) + L 1 ±Vtan 2 x + l \ tan 2 a; Ex. 2. Find cos a; in terms of each of the other five functions. (a) ° 0SX = iic^' from (28) (b) cosx = ± Vl - sin 2 x, from (29) 1 (c) cosx = substitute (36) in (a) ±Vl + tan 2 x (d) cos x = ± A /l- 1 = ±Vcsc2a ~ 1 , substitute (26) in (b) \ CSC 2 X CSC X , . 1 cotx (e) cosx = _____ = • Substitute (30) m (c) ± I 1 ±Vcot 2 x + l \ cot 2 x Ex. 3. Find tan x in terms of each of the other five functions. (a) tanx = — — , from (30) cotx tt>) tan x = ± Vsec 2 x - 1, from (31) (c) tanx = - . from (32) ±Vl — sin 2 x ± Vl - cos 2 (d) tanx = , from (32) COS X (e) tanx = ■ Substituting (38) in (a) ± Vcsc 2 x — 1 Ex. 4. Prove that sec x — tan x • sin x = cos x. Solution. Let us take the first member and reduce it by means of the formu- las (26) to (39), p. 60, until it becomes identical with the second member. Thus 1 sinx . sec x — tan x ■ sin x = sin x cos x cos x [Since sec x = and tan x = I cos x cos x \ 1 — sin 2 x cos 2 x cos x cos x = cosx. Ans. (23), p. 50 * Usually it is best to change the given expression into one containing sines and cosines only, and then change this into the required form. Any operation is admissible that does not change the value of the expression. Use radidals only when unavoidable. 62 PLANE TRIGONOMETRY Ex. 5. Prove that sin x (sec x + osc x) — cos x (sec x — esc x) = sec x esc x. Solution, sin x (sec x + esc x) — cos x (sec x — esc x) [" = sinx! 1 1 — cosxl ; ) \cosx sun/ \cosx sin a;/ sinz J Since secx= and osc a: cosx sinx cosx cosx sinx sin x cos x cos x sin x sin 2 x + cos 2 x cos x sin x _1 1 cos x sin x cos x sin x = sec x esc x. 4ns, (23), p. 59 EXAMPLES 1. Find sec x in terms of each of the other five functions of x. Ans. 1 - 1 ± Vl + tan 2 x , - oosx ±Vl-sin 2 x ± Vcof 2 X + 1 cscx cotx ±Vcsc 2 x-l 2. Find cot x in terms of each of the other five functions of x. 1 / — : r ±Vl-sin 2 x Ans. , ±vcsc 2 x — 1, : 1 cosx tanx " sinx ±Vl — cos 2 x ±Vsec 2 x- 3. Find esc x in terms of each of the other five functions of x. Ans. - , ± Vl + cot 2 X , — sinx ±Vl- ±Vtan 2 x + l secx cos 2 x tan x ± Vsec 2 x — 1 4. Prove the following : (a) cos x tan x = sin x. (b) sin x sec x = tan x. (c) sin y cot y = cos y. (d) (1 + tan 2 y) cos 2 y = 1. (e) sin 2 A + sin 2 4 tan 2 A = tan 2 4. (f ) cot 2 4 - cos 2 A = cot 2 4 cos 2 4 . (g) tan A + cot.4 = sec A esc A. (h) cos 4. esc A = cot 4. (i) cos 2 A — sin 2 A = 1-2 sin* A. (j) cos 2 4 - sin* A = 2 cos 2 A - 1. (k) (l + cot 2 B)sin 2 .B = l. (1) (csc 2 ^ - 1) sin 2 A = cos 2 A. (m) see 2 A + cso 2 ^. = sec 2 JL esc 2 A. (n) cos 4 C - sin* C + 1 = 2 cos 2 C. (o) (sinx + cosx) 2 + (sinx — cosx) 2 = 2. (p) sin 8 x cos x -4- cos 3 x sin x = sin x cos x. (q) sin 2 B + tan 2 B = sec 2 B — cos 2 #. sin i/ (r) coty + = esc y. 1 4- cos j/ (s) cos B tan B + sin B cot B = sin B + cos B. (t) see x cso x (cos 2 x — sin 2 x) = cot x — tan x. . . cosC sinC . _ _ (u) - — - — - + — = sin C + cosC. 1 — tan C 1 — cot C sin z 1 + cos 2 (v) 1- -;—. = 2 cscz. 1 + cos z sin z CHAPTEE IV TRIGONOMETRIC ANALYSIS 32. Functions of the sum and of the difference of two angles. We now proceed to express the trigonometric functions of the sum and dif- ference of two angles in terms of the trigonometric functions of the angles themselves.* The fundamental formulas to be derived are : (40) sin (x + y) = sin x cos y -+- cos x sin y. (41) sin (x — y) = sin x cos y — cos x sin y. (42) cos (x + y) = cos x cos y — sin x sin y. (43) cos (x — y) — cos x cos y + sin x sin y. 33. Sine and cosine of the sum of two angles. Proofs of formulas (40) and (42). Let the angles x and y be each a positive angle less than 90°. In the unit circle whose center is 0, lay off the angle AOP = x and the angle POQ = y. Then the angle A OQ = x + y. O O E C O E In the first figure the angle x + y is less than 90°, in the second greater than 90°. Draw QC perpendicular to OA. Then (a) sin (x + y) = CQ, and (b) cos (x + y) = OC. Draw QD perpendicular to OP. Then (c) sw.y=DQ, and (d) cos y = OD.f * Since x and y are angles, their sum x + y and their difference x - y are also angles. Thus if x = 61° and y = 23°, then x + y =84° and x - y = 38°. The student should observe that sin (x + y) is not the san.e as sin x + sin y, or cos(x - y) the same as cos x - cos y, etc. t The student will see this at once if the book is turned until OP appears horizontal. 63 64 PLANE TRIGONOMETRY Draw DE perpendicular and DF parallel to OA. Then angle DQF = angle AOP (= x), having their sides perpendicular each to each. From (a), (e) sin (x + y) = CQ = CF + FQ = ED + FQ. ED being one side of the right triangle OED, we have ED = OD • sin x. from (7), p. 11 But from (d), OD = cos y. Therefore (f ) ED = sin x cos y. FQ being one side of the right triangle QFD, we have FQ = DQ- cos x. from (8), p. 11 But from (c), DQ = sin y. Therefore (g) FQ = cos x sin y. Substituting (f) and (g) in (e), we get (40) sin(x + y) = sin x cosy + cos x sin y. To derive (42) we use the same figures. Erom (b), (h) cos (x + y) = OC = OE -CE = OE - FD* OE being one side of the right triangle OED, we have OE = OD cos x. from (8), p. 11 But from (d), OD =-. cos y. Therefore (i) OE = cos x cos y. FD being a side of the right triangle QFD, we have FD = DQ sin x. from (7), p. 11 But from (c), DQ = sin y. Therefore (j) FD = sin x sin y. Substituting (i) and (j) in (h), we get (42) cos (* + y) = cos jr cosy — sinx siny. In deriving formulas (40) and (42) we assumed that each of the angles x and y were positive and less than 90°. It is a fact, however, that these formulas hold true for values of x and y of any magnitude whatever, positive or negative. The work which follows will illus- trate how this may be shown for any particular case. * When x + y is greater than 90°, OC is negative. TRIGONOMETRIC ANALYSIS 65 Show that (42) is true when x is a positive angle in the second quadrant and y a positive angle in the fourth quadrant. Proof. Let x = 90° + x' and y = 270° + y'*; then x + y = 360° + (x' + y') and (k) x' = x - 90°, y' = y- 270°, x'+y' = x + y- 360°. cos (x + y) = cos [360° + (x' + y^] = cos (x' + y0 by § 29, p. 57 = cos x' cos y' — sin x' sin y' by (42) = cos (x - 90°) cos (y - 270°) - sin (x - 90°) Sin (y - 270°) from (k) = sinx(— siny) — (— cosxcosy) by §29, p. 57 = cos x cos y — sin x sin y. Q. e. d. Show that (40) is true when x is a positive angle in the first quadrant and y a negative angle in the second quadrant. Proof. Let x = 90° - x' and y = - 180° - y'; then x + y = - 90° - (x' + y') and (1) x'=90°-x, y' = -180°-y, s' + y' = -90°-(x + y). sin(x + y) = sin [- 90° - (x' + y^] = - cos(x' + y') by § 28, p. 56 = — [cos x' cos y' — sin x' sin y 7 ] by (40) = -[cos(90°-x)cos(-180°-y)-sin(90°-x)sin(-180°-y)] from(l) = — [sinx(— cosy) -- cosxsiny] by §29, p. 57 = sin x cos y + cos x sin y. q. e.d. EXAMPLES 1. Find sin 75°, using (40) and the functions of 45° and 30°. Solution. Since 75° = 45° + 30°, we get from (40) sin 75° = sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30° 1 V5 1 1 . K = from p. 5 V2 2 V2 2 V3 + 1 . = Ans. 2V2 2. Find cos(x + y), having given sinx = f and einy = -fg,x and y being positive acute angles. Solution. By the method shown on p. 30 we get first sin x = f , cos x = f , sin y = -^, cosy = |§. Substituting these values in (42), we get aoB{x + V ) = t-ii-i- T % = ft- Ans. * The student should note that x" and y' are acute angles. 66 PLANE TRIGONOMETRY 3. Show that cos 75° = — , using the functions of 46° and 30°. 2V2 4. Prove that sin 90° = 1 and cos 90° = 0, using the functions of 60° and 30°. 5. If tan x = I and tan y = ^ ¥ , find sin (x + y) and cos (x + y) when x and y are acute angles. Ans. sin (x + y) = £ , cos (x + y) = f . 6. By means of (40) and (42) express the sine and cosine of 90° + x, 180° + x, 270° + x, in terms of sin a; and cosx. 7. Verify the following : (a) sin(45° + s) = C ° S ^, Sina; . (c) sin (y + 3 Q°) = ^ sin ^ + C0S ^ ■ (0)^(60°+^=°°^-^^. (d)co S( B + 45° ) = COS ^_ sinJ? . 8. Find sin (A + B) and cos (A + B), having given sin A = \ and sin B = %. + V5 + 2V3 -tVl5 — 2 Ans. sm(A + B) = ± ± , cos(A + B) = ±-^ — 1 34. Sine and cosine of the difference of two angles. Proofs of formulas (41) and (43). It was shown in the last section that (40) and (42) hold true for values of x and y of any magnitude whatever, positive or negative. Hence (41) and (43) are merely special cases of (40) and (42) respectively. Thus, from (40), sin (x + y) = sin x cos y + cos x sin y. Now replace y by — y. This gives (a) sin (x — y) = sin x cos (— y) + cos x sin (— y). But cos (— y) = cos y, and sin (— y) = — 6in y. from p. 55 Substituting back in (a), we get (41) sin (x — y) = sin x cos y — cos x sin y. Similarly, from (42), cos (x + y) = cos x cos y — sin x sin y. Now replace y by — y. This gives (b) cos (x — y) = cos x cos (— y) — sin x sin (— y). But cos (— y) = cos y, and sin (— y) = — sin y. from p. 55 Substituting back in (b), we get (43) cos (x — y) = cos xcosy + sin * sin y. TRIGONOMETRIC ANALYSIS 67 EXAMPLES 1. Find cos 15°, using (43) and the functions of 45° and 30°. Solution. Since 15° = 45° — 30°, we get from (43) cos 15° = cos (45° - 30°) = cos 45° cos 30° + sin 45° sin 30° _j_ V£ j_ i _ V2"' 2 V2'2 _ Vi + i — 7=- - Ans. 2V2 Work out the above example, taking 15° = 60° — 45°. 2. Prove sin (60° + x) — sin (60° - x) = sin x. Solution. sin (60° + x) = sin 60° cos x + cos 60° sin x. by (40) sin (60° - x) = sin 60° cos x - cos 60° sin x. by (41 ) .-. sin (60° + x) — sin (60° — x) = 2 cos 60° sin x by subtraction = 2 • J ■ sin x = sin x. .4ns. V§ — 1 3. Show that sin 15° = -, using the functions of 45° and 30°. 2V2 ■ 4. Find sin(x — y) and cos(x — y), having given sinx = ^ and sin?/ = -^, x and y being acute angles. . . . , 2V2-V15 , , 2V30 + 1 Ans. sm(x-y) = — , cos(x-j/) = — 5. Find sin(x — y) and cos(x — y), having given tanx = ^ and tany = J, x and y being acute angles. Ans. sin (x — y) = fci cos ( x — V) = If- 6. By means of (41) and (43) express the sine and cosine of 90° — x, 180° — x, 270° — x, 360° — x, in terms of sinx and cosx. 7. Verify the following : (a) sin(45° - x) = ™*^* . (c) sin( V - 30°) = ^^~ C0Sy . (b) CO s(60°-^)= C °^ 2+ 2 V5sin ^. W cos(i ? + 45° ) = £^^. (e) sin (60° + x) - sin x = sin (60° - x). (f ) cos (30° + y) - cos (30° -y) = - sin y. (g) cos (45° + x) + cos (45° — x) = V2 cos x. (h) sin (45° + P) - sin (45° - P) = V2 sinP. (i) cos (Q + 45°) + sin (Q - 45°) = 0. (j) sin (x + y) sin (x — y) — sin 2 x — sin 2 y. (k) sin (x + 2/ + z) = sinx cos j/ cos z + cos x sin 2/ cos z + cos x cos y sin z — sin x sin y sin z. Hint, sin (x + y + s) -= sin (a; + y) cos 2 + cos (x + y) sin z. 68 PLANE TRIGONOMETRY 35. Tangent and cotangent of the sum and of the difference of two angles. From (22), p. 59, and (40) and (42), p. 63, we get sin (x + y) sin x cos y + cos x sin y \e\W ( 3T I iy) ~~ ■ — — — • ^ y ' cos (a; + y) cos x cos y — sin x sin y Now divide both numerator and denominator by cos x cos y. This sin x cos y cos x sin y cos x cos y cos x cos y S lves sin x cos y cos x sin y tan (x + y) = cos x cos y sin a; sin y cos X cos ^ cos x cos y sin x siny cos x cos y sin a; sin y cos x cos y sin a: , siny Since = tan x and = tan y, we get cos a; cosy ,..-. , tanx + tany (44) tan (* + ?) = ■ 1 — tan x tan y In the same way, from (41) and (43) we get tanjr — tany (45) tan (* — y) = — — v ' v ' 1 + tan x tan y From (22), p. 59, and (40) and (42), p. 63, we get cos (x + y) cos x cos y — sin x sin y cot (a; + y) = . ) J ! = r-^- v 7 sin (a; + y) sin a; cos y + cos x sin y Now divide both numerator and denominator by sin x sin y. This sdves ° cos a; cosy sin a: sin y , , . sin x sin w sin x sin w cot (x + y) = -, 2 __£ v ' sin a; cos y cos a; sin y sin x sin y sin a; sin y cos aj cos y -. :— ^ — 1 sin x sin y cos y cos x sm y sin x . cosx cosy Since — — • = cot x, and — — - = cot y, we get smx smy , ABS ., , N cotjrcoty — 1 (46) cot (x + y) = — v ' K ' cot y + cot x TEIGONOMETKIC ANALYSIS 69 In the same way, from (41) and (43) we get ,._. L , . cot x cot y + 1 (47) cot (x — y) = 1-!— . v ' v ' cot y — cot x Formulas (40) to (47) may be written in a more compact form as ' "' "" ' ' sin (x ± y) = sin x cos y ± cos x sin y, cos (x ±y) = cos x cos y zp sin x sin y, , . . tan x ± tan y tan(a; ± y) = ■ — =^-i v ' 1 zp tan x tan y . . . . cot x cot «il cot (x ± y) = r - ^ — ^ ' cot jr ± cot x The formulas derived in this chapter demonstrate the Addition Theorem for trigonometric functions, namely, that any function of the algebraic sum of two angles is expressible in terms of the func- tions of those angles. EXAMPLES 1. Find tan 16°, using (45) and the functions of 60° and 45°. Solution. Since 15° = 60° - 45°, we get from (45) tanl5° = tan(60°-45°) = ^an 60° - tan 45° = Vg-1 = g _ ^ ^ 1 + tan 60° tan 45° i + V§ Work out above example, taking 15° = 45° — 30°. 2. Find tan (a; + y) and tan(x — y), having given tanx = \ and tany = \. Ans. tan (x + y) = fy, tan (x — y) = f . 3. Find tan 75° from the functions of 45° and 30°. Ans. 2 + V& 4. Verify the following : 1 + tanx sin(x + y) tanx + tan?/ (a) tan (45° + x) = (f ) -^— = 1 — tan x sm (x — y) tan x — tan y (h)'cot (y -45°) = i±^. <*) tanx + tan, = ^|±^. (c) tan( 4 -60°) = ^^- W ^ " COt * = SfsS " 1 + V3tan^ cos(x + y) V3cotB-l (i) 1- tanxtany = * '-. (d) cot (P + 30°) = VdCOti * -i • oos * cos V cotB + V3 ... . D ,„ n cos(P+Q) (l) cot P cot — 1 = ! — • (e) tan (x ±45°) + cot (xt 45°) = 0. w/ ^ sin P sin Q 36. Functions of twice an angle in terms of the functions of the angle. Formulas (40) to (47) were shown to hold true for all possible values of x and y ; hence they must hold true when x equals y. To find sin 2 x we take (40), sin (x + y) = sin x cos y + cos x sin y. 70 PLANE TRIGONOMETRY Replace y by x. This gives sin (x + x) = sin x cos x + cos x sin a;, or (48) sin 2 Jf = 2 sin x cos jr. To find cos 2 a; we take (42), cos (x + y) = cos a; cos y — sin a; sin y. Replace y by a;. This gives cos (x + x) = cos a; cos a; — sin a; sin x, or (49) cos 2 x = cos 2 x — sin 2 *. Since cos 2 x = 1 — sin 2 x, (49) may be written (49 a) cos 2 x — 1 — 2 sin 2 jr. Or, since sin 2 x = 1 — cos 2 x, (49) may also be written (49b) cos2* = 2 cos 2 * — 1. To find tan 2 x we take (44), _ tan x + tan y *■ ^' 1 — tan a; tan y Replace y by x. This gives . tan x + tan a; tan ( x + x) = ■ ; > or ^ ' 1 — tan x tan x 2 tan jc ( 5 °) Un2x =T-^Fx- 37. Functions of multiple angles. The method of the last section may readily be extended to finding the functions of nx in terms of the functions of x. To find sin 3 x in terms of sin x we take (40), sin (x + y) = sin x cos y + cos x sin y. Replace y by 2 x. This gives sin (x + 2 x) = sin x cos 2 x + cos x sin 2 x, or sin 3 x = sin x (cos 2 x — sin 2 x) + cos x (2 sin x cos x) by (49), (48) = 3 sinx cos 2 x — sin 8 x = 3sinx(l— sin 2 x)— sih'x by (23), p. 5& = 3 sin x — 4 sin 8 x. Ans. To rind tan 4 x in terms of tan x, we take (44), (50), ,_ _ . 2 tan 2 x 4 tan x (1 — tan 2 x) . tan4x=tan(2x + 2x) = 1 _ t?in22a! = 1 _ 6tan2a; + tan 4. Ans. V > TBIGONOMETBIC ANALYSIS 71 EXAMPLES 2 1. Given sinx = — — , x lying in the second quadrant; find sin 2x, cos 2x, tan 2 x. "V6 2 Solution. Since sin x = — and x lies in the second quadrant, we get, using method on p. 30, "^ sinx=- — , cosx = , tanx = — 2. VE V6 Substituting in (48), w6 get 4 2 / 1 \ sin2x = 2smxcosx = 2 ( — ) = — VE\ VEJ 5 Similarly, we get cos2x =— § by substituting in (49), and tan 2x = £ by substituting in (50). 2. Given tan x = 2, x lying in the third quadrant ; find sin 2 x, cos 2 x, tan 2 x. -4ns. sin2x= f, cos2x = — §, tan2x = — $. 3. Given tan x = - ; find sin 2 x, cos 2 x, tan 2 x. 6 2a6 6 2 -a 2 2 aft .4ns. ± a 2 + 6 2 6 2 + a 2 6 2 - a 2 4. Show that cos3x = 4 cos 8 x — 3 cosx. * ou , w . o .i 3 tan .4- tan 8 ^4 5. Show that tan 3 A = 1-3 tan 2 ^i 6. Show that sin 4 x = 4 sin x cos x — 8 sin 8 x cos x. 7. Show that cos4P = 1-8 sin 2 £ + 8 sin* P. 8. Show that sin 5x = 5 sinx — 20 sin 8 x + 16 sin 6 x. 9. Show that tan (45° + x) — tan (45° - x) = 2 tan 2 x. 10. Show that tan (45° + G) + tan (45° - C) = 2 sec 2 C. 11. Verify the following : . „ 2tanx ._ „ csc 2 x (a) sm2x = — • (f) sec2x = — -• w 1 + tan 2 x csc 2 x - 2 1 — tan 2 x (g) 2 esc 2 a = sec s esc s. (b) cos2x= l + tan2a .- (h 1 ) cot* -tan* = 2 cot 2*. (c) tanP + cotP = 2 esc 2P. (i) cos 2x = 2_-sec 2 X - (d) cos2x = cos 4 x - sin 4 x. sec 2 x (e) (sinx + cos x) 2 = 1 + sin 2 x. (j) (sin x - cos x) 2 = 1 - sin 2 x. 12. In a right triangle, C being the right angle, prove the following -. OB sin 2 .i-sin 2 P „. tan2 ,_ 2ab , (b) sin( J 4-P) + eos24 = 0. (i) sin 2 .4 = sin 2 P. (c) sin(4-P) + sin(24 + C) = 0. (j)sin2X=^. (d) (sin4 - sinP) 2 + (cos4 + cosP) 2 = 2. c 2 (e ) J«±I + A /3= 4===L- (k)cos24= 62 - a2 (f ) tan £ = cot A + cos C. n\ gm 3 4 = (g) cos 2 4 + cos 2 P = 0. c a 3 oW - a 8 72 PLANE TRIGONOMETRY 38. Functions of an angle in terms of functions of half the angle. From (48), p. 70, v ' sin 2 a; = 2 sm a; cos «. Eeplaee 2xby x, or, what amounts to the same thing, replace * by ^ • This gives (51) sin * = 2 sin - cos - • From (49), p. 70, cos 2 a; = cos 2 x — sin 2 x. Replace 2 x by x, or, what amounts to the same thing, replace x by ^ • This gives X X (52) cos x = cos 2 sin 3 - • ft & From (50), p. 70, tan 2 a; = 1 _ t ^ a . • Replace 2 x by a;, or, what amounts to the same thing, replace x X — ■ This gives 2tan- (53) tanx = . 1 _ tan 2 £ 2 39. Functions of half an angle in terms of the cosine of the angle. From (49 a) and (49 b), p. 70, we get 2 sin 2 x = 1 — cos 2 x, and 2 cos 2 a: = 1 + cos 2 x. Solving for sin x and cos x, sin a; = ± ll— cos 2 a: N — 5 -±>j£ i j. + cos 2 a; and cos x = ± -\l Replace 2 x by x, or, what amounts to the same thing, replace x by - • This gives -~ % , ,_., . x 1 1 — cos x (54) sil--=± A J - , (55) cos|.c±-y|- + cos* \ TEIGONOMETRIC ANALYSIS To get tan - we divide (54) by (55). This gives 73 . x sm- , x 2 tan- = -_ cos- N ■ cos a; -J- + eosa; i) or, (56) tan - = db \\- — 2 \l + COS X + COS X Multiplyin g both nu merator and denominator of the right-hand member by Vl + cos x, we get * . V"" ( Verify t sin 82° * \ x sin x tan- = ; 2 1 + cos x or, muj.Jjjiying both numerator and denominator by Vl — cos x, we get (58) X 1 — cos X tan — = . 2 sinx Since tangent and cotangent are reciprocal functions, we have at once from (56), (57), and (58), (59) (60) (61) 40. Sums and differences of functions. From p. 63, (40) sin (x + y) = sin x cos y + cos x sin y. (41) sin (x — y) = sin x cos y — cos x sin y. (42) cos (x + y) = cos x cos y — sin x sin y. (43) cos (x — y) = cos x cos y + sin a; sin y. X cot— — ± N 1 + cos X 2 1 — cos X X cot— = 2 1 + COS X sinx X cot- — 2 sinx 1-c osx t -\ / l — COS3? VTT cos x Vl— co&ne sin x -cos a; 1 + cosa; I+coe-r The positive sign vr> ly of the radical is taken since 1 + cos x can never be negative and tan — and sin x alwajgo a-ve like signs. 74 PLANE TRIGONOMETRY First add and then subtract (40) and (41). Similarly, (42) and (4$.'^ This gives (a) sin (x + y) + sin (a; — y) = 2sina;cosy. Adding(40)and(41) (b) sin (x + y) — sin (x — y)= 2 cos x sin y. Subtracting (4 l)from (40) (c) cos(x + y) + cos(x — y)= 2 cos x cosy. Adding (42) and (43) (d) cos(sc + y) — cos(a; — y)=— 2sina;siny. Subtracting(43)from(42) Let x + y =A x + y = A and x — y = B x — y =B Adding, 2x=A+B Subtracting, 2 y = A — B x = $(A+B). y = i(A-B). Now replacing the values of x + y, x — y, x, y in tern/ ' f A and B in (a) to (d) inclusive, we get I (62) sin A + sin B = 2 sin | (4 + B) cos \(A — B)* (63) sinX — sinB= 2 cos|(A + B) sin §(.4 — B). (64) cos^ + cosB= 2cos|(A + B)v)s\{A — B). (65) cos.4 — cos.B = — 2 &i.n.\(A + B) sin§(.4 — B). Dividing (62) by (63), member for member, we obtain sin^+sinB _ 2 sin £ (A +B) cos J {A —B) siaA — sin.B — 2 cos £ (A + B) sin %(A—B) _sinK£-KB) cos^(^--B) But (66) cot^(A-B) oos^(A+B) sin %(A—B) = tan £ (4 + J3) cot i (A — B). 1 tan ±(A—B) Hence sin .4 + sinB _tanl(A + B) amA — sinB ~~ tan|(4 — B) EXAMPLES 1 1. Find sin 22 £°, having given cos 45° = VI * /r Solution. From (64), sin- = ± -» / — cosx Let x = 45°, then - = 22 £°, and we , 11-- sin 22£° = V2 1 \ = !V2~ 2 4ns. TEIGONOMETEIG ANALYSIS 75 2. Reduce the sum sin 7 x + sin 3 a; to the form of a product. Solution. From (62), sin^. + sinB = 2sin^(^ + B)cos{-(A-B). Let4 = 7x, B = 3x. Then A + B = lOx, and^.-.B = 4x. Substituting back, we get sin7a; + sin3x = 2sm5xcos2x. Ans. 3. Find cosine, tangent, and cosecant of 22j . Xna. - V2+V2, Vis - 1, 2 Va'WI 4. Find sine, cosine, and tangent of 15°, having given cos 30° = Ans. - V2-V3; - V2+V3"; -xl— 2 2 \| 2 + V3 5. Verify the following: 2 2 >2+V3 (a) sin 82° + sin 28° = cos 2°. ,\(- x X V_ 1 (b) sin50° + sinl0° = 2sin30°cos20°. (S) \ Sm 2~ COS 2) - 1 ~ smx - (c) cos 80° - cos 20°= - 2 sin 60° sin 30°. x sin £ x (d) cos 5x + cos9x = 2 cos7xcos2x. ' ' an 4 — 1 + cos 1 x ' ,. sin7x — sin5x „ sinl-r ( e ) — ; T- = tanx - (i\ cot - - * . 0087^ + 00850; (,C °4 1-cos^x sin 33° + sin 3° _ sin4 + sin _g , (f) cos33° + cos3°- tan18 - 0) cos ^ +cosB = to 2^ + B) - a; 1 6. Find sine, cosine, and tangent of -, if cosx = -and x lies in the first quadrant. V3 Vfi V2 4m - IT IT 1" 7. Find sine, cosine, and tangent of - , if cos x = a. Il — a k + a /l-o ^ s - ± V^- ,± V^-' ± Vi+^- 8. Express sine, cosine, and tangent of 3 x in terms of cos 6 x. /l — cos 6 x Vl + cos 6 x /l — cos6x Ans . ±> /— ± , iVl + cosex' 9. In a right triangle, being the right angle and c the hypotenuse, prove the following : „B c — a ... a-b , A—B (a) sin"- = - — -■ (d) - — - = tan w 2 2c a + b (b) (cos I + sin I J= £±i • (e) tan | = w 2 2c 41. Trigonometric identities. A trigonometric identity is an expres- sion which states in the form of an equation a relation which holds true for all values of the angles involved. Thus, formulas (26) to (39), p. 60, are all trigonometric identities, since they hold true for all 76 PLANE TRIGONOMETRY values of x; also formulas (62) to (66), p. 74, are identities, since they hold true for all values of A and B. In fact, a large part of the work of this chapter has consisted in learning how to prove identities by- reducing one member to the form of the other, using any known identities (as in Ex. 4, p. 61). Another method for proving an identity is to reduce both members simultaneously, step by step, using known identities, until both mem- bers are identical in form. No general method can be given that will be the best to follow in all cases, but the following general directions will be found useful. General directions for proving a trigonometric identity.* First step. If multiple angles, fractional angles, or the sums or differences of angles are involved, reduce all to functions of single angles only f and simplify. Second step. If the resulting members are not readily reducible to the same form, change all the functions into sines and cosines. Third step. Clear of fractions and radicals. Fourth step. Change all the functions to a single function. In case the second step has been taken, this means that we change to sines only or to cosines only. The two members may now easily be reduced to the same form. Ex. 1. Prove the identity 1 + tan 2 x tan x = seo 2 x. Solution. Since tan 2 x = and sec 2 x = = , the equation becomes : l-tan»x cos2x sec*x-sm 2 x First step. 1 + 1 — tan 2 x cos 2 x — sin 2 x l + tan 2 x_ 1 1 — tan 2 x cos 2 x — sin 2 x sin 2 a; 1 + cos 2 a 1 sin 2 x cos 2 x — sin 2 x cos 2 x cos 2 x + sin 2 x 1 or, simplifying, Second step. or, simplifying, cos 2 x — sin 2 x cos 2 x — sin 2 x Third step. cos 2 x + sin 2 x = 1, ° r 1 = 1. from (23), p. 59 * In -working out examples under this head it will appear that it is not necessary to take all of the steps in every case, nor will it always be round the best plan to take the steps in the order indicated. t For instance, replace sin 2 a: by 2 sin a: cos x, tan 2 a; by - — anx , cos (a; + y) by cos x cosy — sinarsiny, etc. 1 — tana; TRIGONOMETRIC ANALYSIS 77 Ex. 2. Prove ^ (x + y) _ tan a + tan y sin (x — y) tan x — tan y Solution. Since sin (x + y) = sin x cos y + cos x sin y, and sin (x — y) = sin x cos y - cos x sin y, the equation becomes : First step. sin x cos y + cos x sin y sin x cos 2/ — cos x sin 2/ tan x + tan y tan x — tan y sin x sin 2/ Second step. sin x cos 2/ + cos x sin 2/ cos x cos 2/ sin x cos j/ — cos x sin j/ sin x sin j/ cos x cos y Simplifying, sin x cos y + cosx sin y sin x cos y + cos x sin y sin x cos ?/ — cos x sin 2/ sin x cos 2/ — cos x sin 2/ or, 1 = :1. EXAMPLES Prove the following identities : 1. tan x sin x + cos x = sec x. 2. cot x — sec x esc x (1 — 2 sin 2 x) = tan x. 3. (tanx +. cot x) sin x cosx = 1. sin 2/ 1 — cos 2/ 4. 1 + cos y sin y -£. , . ■ smA . . . 5. -\/ = sec4 — tan.4. . + sin^i 6. tan x sin x cos x + sin x cos x cot x = 1. 7. cot 2 x = cos 2 x + (cot x cosx) 2 . 8. (sec 2/ + osc y) (1 — cot 2/) = (sec y — esc y) (1 + cot 2/). 9. sin 2 z tan z + cos 2 z cot z + 2 sin z cos z = tan z + cot z. 10. sin 6 x + cos 6 x = sin 4 x + cos 4 x — sin 2 x cos 2 x. 11. sinBtan 2 .B + csc.Bsec 2 .B = 2tanJBsec£+ cscB — sinB. 12. Work out (a) to (v) under Ex. 4, p. 62, following the above general directions. 13. cos(x + 2/)cos(x — y) = cos 2 x — sin 2 y. 14. sin (A + B) sin (A - B) = cos 2 B - cos 2 A. 15. sin (x + y) cos y — cos (x + 2/) sin y = sin x. cos (x — y) _ 1 + tan x tan y 16. cos (x + y) 1 — tan x tan y sin (4 - .B) 17. tan^A — tanB = 18. cotx + cot 2/ = cos J. cos I? sin (x + 2/) sin x sin 2/ 19. sin x cos (2/ + z) — sin 2/ cos (x + z) = sin (x - y) cos z. 78 PLANE TRIGONOMETRY 20. *"■('- »> + *■"*»= t»n g. . 1 — tan (0 — (p) tan cj> 21. cos (x — y + z) = cos x cos y cos z + cos x sin y sin z — sin x cos y sin z + sin x sin ?/ cos z. 22. sin (x + y — z) + sin (x + z — y) + sin (y + z — x) = sin (x + y + z) + 4 sin x sin y sin z. 23. cos x sin (y — z) + cos y sin (z — x) + cos z sin (x — y)-= 0. 24. Work out (a) to (k) under Ex. 7, p. 67, and (a) to (j), under Ex. 4, p. 69, following the above general directions. 25 l + sin2x = Aanx + iy ^ ^ = l-sin2X 1 — sin 2 x \tan x — 1/ cos 2 .4 sin2x „. cos 8 x + sin 8 x 2 — sin 2 x 26. tan x = 1 + cos 2 x 30. 27. sin2x 31. 1 — cos 2 x 28. cot.4 — 1 /l — sin 2A cot4 + l~ Vl + sin2vi 32. 33. sin 3 x = 4 sin x sin (60° + x) sin (60° -■). 34. sin 4 x = 2 cos 2 x. sin2x 35. sin 4 .B = 4 cos 2 B sin B cos B. cos x + sin x 2 sin 3 x — sin x cos 3 x + cos x sin 3 x — sin x cos x — cos 3 x = tan x. cot 2 x. 36. Work out (a) to (j) under Ex. 11, p. 71, following the above general directions. 37. sin 9x — sin7x = 2 cos8xsinx. 1-tan 2 ? 38. cos7x + cos5x= 2cos6x cosx. .„ 2 46. = cosx. 39. sin5x-sin2x =eot 7x 1+tan 2 * = uui • • o cos 2 x — cos 5x2 z acx I ■ x , X V i , • 2 tan - 40. sin - + cos - = 1 + sin x. 2 \ 2 2/ 47. — = sinx. ., 1+seoy „ „y 1 + tan 2 - 41. — ^ = 2cos 2 ^. 2 sec y 2 x 48. 1 + tan x tan - = sec x. ._ sin ^.+ sin B 1, , _ 2 42. -Z - = -cot-(4-B). cos^-cosB 2 X 49. .tan? + 2sin^cotx = sinx. 2 2 43. 1 + tan - cos 9 2 60. 1 + cot? - = . 1 ~ Sine 1-tan*' 2 Slna:tan l 2 xx 44. cos3or — cos7ar=2sin5arsin2cr. tan 2 - + cot 2 - t K . , x 51. 2 2 l"+cos 2 x 45. cot - + tan - = 2 esc x. tan 2 - - cot 2 - 2 cos x 2 2 2 2 52. Work out (a) to (f) under Ex. 5, p. 76, following the above general directions. CHAPTEE V GENERAL VALUES OF ANGLES. INVERSE TRIGONOMETRIC FUNC- TIONS. TRIGONOMETRIC EQUATIONS 42. General value of an angle. Since all angles having the same initial and terminal sides have the same functions, it follows that we can add 2 v to the angle or subtract 2 it from the angle as many times as we please without changing the value of any function. Hence all functions of the angle x equal the corresponding func- tions of the angle 2 mr + x, where n is zero or any positive or negative integer. The general value of an angle having a given trigonometric func- tion is the expression or formula that includes all angles having this trigonometric function. Such general values will now be derived for all the trigonometric functions. 43. General value for all angles having the same sine or the same co- secant. Let x be the leas't positive angle whose sine has the given value a, and consider first the case when a is positive. Construct the angle x (= XOP), as on p. 31, and also the angle it — x (= XOP 1 ), having the same value a for its sine. Then every angle whose terminal side is either OP or OP' has its sine equal to a, and it is evident that all such angles are found by adding even multiples of 7r to, or subtract- ing them from, x and it — x. Let n denote zero, or any positive or negative integer. When n is even, mr + x includes all the angles, and only those, which have the same initial and terminal sides as x (= XOP). Therefore, when n is even, (.4) mr + x = mr + (- l)"x* * The factor (- 1)» is positive for all even values of n and negative for all odd values of n. 79 80 PLANE TRIGONOMETRY- Again, when n is odd, n — 1 is even, and (n — 1) 7r + (nr — x) in- cludes all the angles, and only those, which have the same initial and terminal sides as it — x (= XOP 1 ). But when n is odd, (*) (n — 1) it + (it — x) = Jwr — a; = mr + (— l) n x. From (4) and (B) it follows that the expression nir +(— l) n a; for all values of n includes all the angles, and only those, which have the same initial and terminal sides as x and 7r — x. In case a is negative, ir — x will be negative, as shown in the figure, but the former line of reasoning will still hold true in every particular. Since sine and cosecant are reciprocal functions, it follows that the expression for all angles having the same cosecant is also nir + (— V) n x. Hence (67) mr + (- l)"x is the general value of all the angles, and only those, which have the same sine or cosecant as x* This result may also be expressed as follows : sin x — sin[n7T + (— l)"Jf]i esc x = csc[n7T + (— I)"*]- o _ Ex. 1. Find the general value of all angles having the same sine as —— Solution. Let x = — in (67). This gives 4 mr + (— 1)" 4 ' Ans. Ex. 2. Find the four least positive angles whose cosecant equals 2. Solution. The least positive angle whose cosecant = 2 is This gives -• Let x = - in (67). 6 6 717T + (— 1)«- * In deriving this rale we have assumed x to be the least positive angle having the given sine. It follows immediately from the discussion, however, that the rule holds true if we replace x Dy an angle of any magnitude whatever, positive or negative, which has the given sine. The same observation applies to the rules derived in the next two sections GENEEAL VALUES OF ANGLES When n = 0, we get 81 *= 30°. When n = 1, we get When n = 2, we get When n = 3, we get 2ir + 150°. : 390°. 3tt--= 510°. Ana. 44. General value for all angles having the same cosine or the same secant. Let x be the least positive angle whose cosine has the given value a, and consider first the case when a is positive. Construct the angle x (=XOP), and also the angle — x (= XOP'), hav- ing the same value a for its cosine. Then every angle whose terminal side is either OP or OP' has its cosine equal to a, and it is evident that all such angles are found by adding even multiples of ir to, or sub- tracting them from, x and — x. Let n denote zero, or any positive or negative integer. Eor any value of n, (A) 2 wrr + x * includes all the angles, and only those, which have the same initial and terminal sides as x (= XOP). Similarly, (B) 2 nir — x includes all the angles, and only those, which have the same initial and terminal sides as -x(=XOP f ). In case a is negative, the same line of reasoning still holds true. Since cosine and secant are reciprocal functions, it follows that the same discussion holds for the secant. Hence, from (A)a,nd(B), (68) 2 nv ± x is the general value of all the angles, and only those, which have the name cosine or secant as x. * in is even and 2» - 1 is odd for all values of n. 82 PLANE TRIGONOMETRY This result may also be expressed as follows : cos x = cos (2 mr d= x), sec x = sec (2 rm ± jc) . Ex. 1. Given aosA = — ; find the general value of A. Also find the five least positive values of A. ~v2 Solution. The least positive angle whose cosine = — is If we let 3*- V2 4 x = — in (68), we get 4 4 = 2nx±^. When n = 0, When ji = 1, When 7i = 2, A 4 = 135°. A = 2tt ± 3?r_ 225° or 495°. A = 4tt ± 3?r_ 4 685° or 855°. Arts. 45. General value for all angles having the same tangent or the same cotangent. Let x be the least positive angle whose tangent has the given value a, and consider first the case when a is positive. Construct the angle x (= XOP), and also the angle it + x (= XOP') having the same value a for its tangent. Then every angle whose terminal side is either OP or OP' has its tangent equal to a, and it is evident that all such angles are found by adding even multiples of ir to, or subtracting them from, x and 7T + x. Let n denote zero, or any positive or negative integer. When n is even, w nir + x includes all the angles, and only those, which have the same initial and terminal sides as x (= XOP). Again, when n is odd, n — 1 is even, and (B) (n — 1)7T +(ir + x)= mr + x includes all the angles, and only those, which have the same initial and terminal sides as ir + x (—XOP 1 ). In case a is negative, the sfime line of reasoning still holds true. GENEKAL VALUES OF ANGLES 83 Since tangent and cotangent are reciprocal functions, it follows that the same discussion holds for the cotangent. Hence, from (A) and (-B), for all values of n, (69) rnr + x is the general value of all the angles, and only those, which have the same tangent or cotangent as x. This result may also be stated as follows: tan* = tan (tot + x), cot x = cot (mr + x). EXAMPLES 1. Given sin A = £ ; find the general value of A. Also find the four least positive values of A. Ans. mr + (- !)»-;, 30°, 150°, 390°, 510°. V§ 2. Given cos^i = — ; find the general value of A. Also find all values of A a n "■ it llr numerically less than 2 it. Ans. i mr ± — ; ±—,± — — 3. Given tan .4 = 1 ; find the general value of A. Also find the values of A numerically less than 4 tt. t jt 3tt bir 7ir 9ir Utt 13tt lbir 44 44 44 44 4 . «. . ~ 1 i ., mr . -v 7r 4. Given sin 2 x = - ; show that s = — - + (— 1)» — . 2 2 12 5. Given cos 3 x ■■ 1 . t . . 2mr , 2tt - ; show that x = ± 2' 3 9 6. In each of the following examples find the general values of the angles, having given Ans. A = mr + (- 1)» ( ± ^j = (a) sin A = ±1. (b) cotx=±Vl$. (c) cosy=±-. (d) tan.B = ±l. (e) cscG = ±-^/2. •■ mr ± 2 x = nir ± 6 (f) sec^=± ■ ,/3 # = mr ± -• O 4 4 A = mr ± — • 84 PLANE TEIGONOMETEY 7. Given sin x = and tan x = ; find the general value of x. 2 V3 Solution. Since sin a; is — and tans is +, x must lie in the third quadrant. The smallest positive angle in the third quadrant which satisfies the condition sin x = — is — , and this angle also satisfies tan x = 2 6 ^ Hence x = 2 wr -\ ■ Ans. 6 „, _Ji. In each of the following examples find all the positive values of x less than 2 ir which satisfy the given equations. (a) cosa; = — — • Ans. —, V2 4 4 .... 1 J 3r 6x 7ir (») sinx = ±— • . V2 4 4 4 4 ir 27r 4t 5tt I'T' T'T' ir 3tt 5-ir Itt T T' T' T 67T 77T T'T" 8'T' w 57r 7w llrr 46. Inverse trigonometric functions. The value of a trigonometric function of an angle depends on the value of the angle ; and con- versely, the value of the angle depends on the value of the function. If the angle is given, the sine of the angle can be found ; if the sine is given, the angle can be expressed. It is often convenient to rep- resent an angle by the value of one of its functions. Thus, instead of saying that an angle is SO , we may say (what amounts to the same thing) that it is the least positive angle whose sine is ^. We then con- sider the angle as a function of its sine, and the angle is said to be an inverse trigonometric function, or anti-trigonometric, or inverse circular function, and is denoted by the symbol sin -11 -, or, arc sin £*, read "inverse sine of \," or, "arc (or angle) whose sine is $." Similarly, cos -1 a; is read "inverse cosine of x," tan" 1 ?/ is read "in- verse tangent of y," etc. If a is the value of the tangent of the angle x, we are now in a position to express the relation between a * Symbol generally used in Continental books. (c) tana = ±V5. (d) COtX: = ±1. (e) COSX : __V3 2 (f) secx : = 2. (g) sin x : 1 = ± 2- INVEKSE TRIGONOMETRIC FUNCTIONS 85 and x in two different ways. Thus, tanx = a, meaning the tangent of the angle x is a; or, x = tan -1 a, meaning x is an angle ivhose tangeni is a. The student should note carefully that in this connec- tion — 1 is not an algebraic exponent, but is merely a part of the mathematical symbol denoting an inverse trigonometric function. tan -1 a does not denote (tana.) -1 = > J tan a but does denote each and every angle whose tangent is a. The trigonometric .functions (ratios) are pure numbers, while the inverse trigonometric functions are measures of angles, expressed in degrees or radians. Consider the expressions tan x = a, x — tan -1 a. In the first we know that for a given value of the angle x, tan x (or a) has a single definite value. In the second we know from (69), p. 83, .that for a given value of the tangent a, tan -1 a (or x) has an infinite number of values. Similarly, for each of the other inverse trigonometric functions. Hence : The trigonometric functions are single valued, and the inverse trigo- nometric functions are many valued. The smallest value numerically of an inverse trigonometric func- tion is called its principal value.* For example, if tan x =1, then the general value of x is, by (69), p. 83, 77" x = tan -1 1 = mr + — > 4 where n denotes zero or any positive or negative integer, and 7 = 45° 4 is the principal value of x. * Hence, if sin x, esc a?, tan x, or cot x is positive, the principal value of x lies between and -^ j if sina:, cscx, tana:, or cota: is negative, the principal value of x lies between and - 1 • 7T 7T If cosx or sec a: is positive, the principal value of x lies between ^ and — -^^ prefer- ence being given to the positive angle. 86 PLAJSTE TRIGONOMETRY Similarly, if cos x = \, then by (68), p. 81, x = cos J ;r = 2 nir ± — > Z o tr where the principal value of x is — = 60°. o Since the sine and cosine of an angle cannot be less than — 1 nor greater than + 1, it follows that the expressions sin -1 a. and cos -1 a have no meaning unless a lies between — 1 and + 1 inclusive. Simi- larly, it is evident that the expressions . sec -1 a and esc -1 a have no meaning for values of a lying between — 1 and + 1. Any relation that has been established between trigonometric functions may be expressed by means of the inverse notation. Thus, we know that cos x = Vl — sin 2 a;. (29), p. 60 This may be written (A) x = cos -1 Vl— sin 2 a;. Placing sin a; = a, then x = sin -1 a, and (A) becomes sin -1 a = cos -1 Vl — a 2 . Similarly, since cos 2 x = 2 cos 2 a; — 1, (49 b), p. 70 we may write (B) 2x = cos- 1 (2 cos 2 x - 1). Placing cos a; = c, then x = cos _1 c, and (B) becomes 2cos- 1 c = cos- 1 (2c 2 -l). Since we know that the co-functions of complementary angles are equal, we get for the principal values of the angles that sin -1 a + cos -1 a = —> < a S 1 tan -1 b + cot- 1 5 = ^> < b sec -1 g + esc -1 c = — • 1 < c INVERSE TRIGONOMETRIC FUNCTIONS 87 We shall now show how to prove identities involving inverse trigonometric functions for the principal values of the angles. Ex. 1. Prove the identity m + n (a) tan _1 m + tan- 1 ?! = tan -1 - 1 — mn Proof. Let (b) ji = tan -1 m and B = tan _1 m. (c) Then tan 4 = m and tan .B = n. Substituting from (b) in first member of (a), we get A + B = tan- 1 — — — , 1 — mn or, what amounts to the same thing, (d) tan(4 + £) = -^±^. 1— mn But from (44), p. 68, . . _. tanA + tanB (e) tanM+B) = w v 1- tan^tanJS Substituting from (c) in second member of (e), we get (f) tan(^ + B) = ^±^. 1 — mn Since (d) and (f) are identical, we have proven (a) to be true. Ex. 2. Prove that (g) sin- 1 ! + cos- 1 !! = sin- 1 ^. Proof. Let (h) .4 = sin -1 ! an< ^ B=cos- 1 ^. (i) Then sin .4 = f and cosB=\%. (j) Also cos 4 = f and sinJB = T 8 T .* Substituting from (h) in first member of (g), we get A + B= sin-i^, or, what amounts to the same thing, (k) sm(A + B) = $l But from (40), p. 63, (1) sin (A + B) = sin A cos B+ cos A sin B. Substituting from (i) and (j) in second member of (1), we get (m) sin(4 + -B) = !-H + W T = i£. Since (k) and (m) are identical, we have proven (g) to be true. The following example illustrates how some equations involving inverse trigonometric functions may be solved. * Found by method explained on p. 30. 88 PLANE TRIGONOMETRY Ex. 3. Solve the following equation for x : tan~ 1 2a + tan _1 3x = — 4 Solution. Take the tangent of both sides of the equation. Thus * tan (tan- 1 2 x + tan -1 3 x) = tan — , tan (tan- 1 2 x) + tan (tan- 1 3 x) , , .... or, — = 1, from (44), p. 68 1 - tan (tan- 1 2 x) tan (tan- 1 3 x) 2x + 3x or, = 1. ' l-2x-3x Clearing of fractions and solving for x, we get x = £ or — 1. x = £ satisfies the equation for the principal values of tan- 1 2 x and tan -1 3x. x = — 1 satisfies the equation for the values tan-!(- 2) = 116.57°, tan- 1 (-3) = -71. 67°. EXAMPLES 1. Express in radians the general values of the following functions : (a) sin- 1 — -■ Ans. nic + (— 1)" — • (e) tan- 1 — = . Ans. mr -\ V2 4 V3 6 (b)dn-i(-^). n»-(-l)-J. (i)Un-i(±Vs). wr±|. (c) cos- 1 — — 2ri7r±^- (g) cot-i(± 1). mr±~. 2 o 4 (d)ooB-i(-i). !w±j- (hJcot-^-LY TC1r + J. 2. Prove the following : (a) tan -1 a — tan- 1 6 = tan- 1 (h) 2 tan- 1 a = tan- 1 l + a& 1-a 2 2a (b) 2tan- 1 a= sin- 1 -■ (i) sin-ia = cos-iVl- a 2 . 1 + a 2 ' (c) 2 sin- 1 a = cos- 1 (1 - 2 a 2 ). (j) sin- i a = tan- 1 Vl-a 2 (d) tan- 1 * = sin- 1 — — ==.. (k) tan- 1 * = cos- 1 Vl + a 2 Vl + a 2 (e) tan- 1 tan- 1 = - . (1) sin- 1 - + sin- 1 — = sin- 1 — . n m + n 4 6 17 85 (f) cos- 1 - + tan- 1 - = tan- 1 -^ (m) cos- 1 - + cos- 1 — = cos- 1 — . 5 5 11- v ' 5 13 65 2 12 119 (g) 2 tan- 1 - = tan- 1 — . (n) tan- 1 - + tan- 1 — = tan- 1 - • 3 5 7 13 9 * The student should remeuiber that tan-' 2 a: and tan- 1 3 a are measures of angles. TKIGONOMETRIC EQUATIONS 89 3. Solve the following equations : (a) tan- 1 ! + tan-!(l - x) = tan-M-). (b) tan-i* + 2 coWx = — . o ... .1- 1 _x + 1 ir (c) tan- 1 [■ tan- 1 = -. x + 2 x+2 4 (d) cos- 1 \- tan- 1 = x 2 + 1 x 2 - 1 3 (e) tan- 1 ^-i^ + tan- 1 ^li = tan- 1 (-7). a: — 1 a; (f) tan- 1 (x + 1) + tan- 1 (x - 1) = tan- 1 — . ol (g) sin -!x + sin- 1 2 x = — . o (h) sm- 1 - + sin- 1 — = - . x x 2 4. Find the values of the following : 2a Ans, 1 x = -• 2 x = V§. X = ± \l x=± V3. x = 2. x = -8, l 14 x = ± 13. (a) sin (tan- 1 — ). \ 12/ Ans. 5 ±13" (d) cos (2 cos -1 a). (b) cot (2 sin- 1 -V 7 ±24- (e) tan (2 tan- 1 a). (c) sin/ tan -1 - + tan~ i> 1 (f) cos(2tan- 1 a). 1-a 2 1 + a 2 ' 47. Trigonometric equations. By these we mean equations involv- ing one or more trigonometric functions of one or more angles. For instance, 2cos 2 x + V3sina; +1 = is a trigonometric equation involving the unknown angle x. We have already worked out many problems in trigonometric equations. Thus, Examples 1-8, pp. 83, 84, are in fact examples requiring the solution of trigonometric equations. To solve a trigonometric equation involving one unknown angle is to find an expression (the student should look up the general value of an angle, p. 85) for all values of the angle which satisfy the given equation. No general method can be given for solving trigonometric equa- tions that would be the best to follow in all cases, but the following general directions (which are similar to those given on p. 76 for proving identities) will be found useful. 90 PLANE TKIGONOMETKY 48. General directions for solving a trigonometric equation.* First step. If multiple angles, fractional angles, or the sums or differences of angles are involved, reduce all to functions of a single anglerf and simplify. Second step. If the resulting expressions are not readily reducible to the same function, change all the functions into sines and cosines. Third step. Clear of fractions and radicals. Fourth step. Change all the functions to a single function. Fifth step. Solve algebraically (by factoring or otherwise) for the one function now occurring in the equation, and express the general value of the angle thus found by (67), (68), or (69). Only such values of the angle which satisfy the given equation are solutions. Ex. 1. Solve the equation cos 2 x sec x + sec x + 1 = 0. Solution. Since cos 2 x = cos 2 x — sin 2 x, we get First step. (cos 2 x — sin 2 x) sec x + sec x + 1 = 0. Second step. Since sec x = 1 this becomes cosx cos 2 x — sin 2 x 1 _ cosx cosx — Third step. cos 2 x — sin 2 x + 1 + cosx = 0. Fourth step. Since sin 2 x = 1 — cos 2 x, we have cos 2 x — 1 + cos 2 x + 1 + cosx = 0, or, 2 cos 2 x + cosx = 0. Fifth step. cos x (2 cos x + 1) = 0. Placing each factor equal to zero, we get cos x = 0, or, from (68), p. 81, x = cos- 1 *) = 2 rwr ± - • Also, 2 cos x + 1 = 0, cosx = — J, ./ 1\ „ , 2tt or, x = cos- 1 1 — -I = 2rnr ± — ■ Hence the general values of the angles which satisfy the equation are 2 nir ± — and 2 nw ± — • £t O The positive angles less than 2 w which satisfy the equation are then ir lir 4ir 37r 2' T' T' T* * In working out examples under this head it will appear that it is not necessary to take all of the steps in every case, nor will it always he found the best plan to take the steps in the order indicated. t For instance, replace cos 2 a; by cos 2 x — sin2 a:, sin(a; + T\ by — v 4 ' vs TEIGONOMETEIC EQUATIONS 91 Ex. 2. Solve the equation 2 sin 2 x + V3 cos x + 1 = 0. Solution. Since sin 2 x = 1 — cos 2 x, we get Fourth step. 2 — 2 cos 2 x + Vs cos x + 1 = 0, or, 2 cos 2 x — Vs cos x — 3 = 0. Fifth step. This is a quadratic in cos x. Solving, we get R V3 cos x = V3 or 2 Since no cosine can be greater than 1, the first result, cosx =V3, cannot be used. From the second result, ■i-f)= 5,r A 2 mr ± — - • Ana. 6 EXAMPLES Solve each of the following equations : 1. sin 2 x = l. Ans. x=mr+( — 1)"( ±— l*=mr± 2. csc 2 x = 2. x = n7r + (-l)»(±-)=mr± 3. tan 2 x = 1. 4. cot 2 x = 3. 5. cos 2 x = -■ 4 4 6. sec 2 x = - - 3 7. 2sin 2 x + 3cosx = 0. 8. cos 2 or — sin 2 a = - ■ 2 9. 2 Vs cos 2 a = sin a. 10. sin 2 y — 2 cos y + - = 0. 4 11. sin.4 + cos.d = V2. 12. 4sec 2 y-7tan 2 # = 3. 13. tanB + cotB = 2. 14. tan 2 x - (1 4 V3) tan x + Vs = 0. » Since the principal value of a: = sin- 1 1 = - and of i = sin- 1 (-!)=- — > X = mr ± — • 4 7T x = mr ± - ■ 6 ■k x = nir ± — • o TT x = mr ± — ■ 6 2tt x = 2 mr ± —— ■ ir a = mr ± — ■ 6 a = mr + (- 1)» 7T I' y = 2 w ± - • o A = 2 rwr + - • 4 7T 2/ = mr ± - ■ 1? = mr H 4 7T IT x = mr + — i mr + .— 4 o 92 PLANE TEIGONOMETEY 15. cot 2 x+( V5 H — )cotx + l=0. Ana. x = mr -\ — -, rnr -\ — -• \ V3/ 6 3 16. tan 2 x + cot 2 x = 2. x =.mr ± - • 4 17. tan / x + -) = 1 + sin 2 x. x = mr, mr — ^ • 18. csex cotx = 2 VI. i = 2mr + - ■ 6 19. sin- = csc a; — cotx. x = 2 mr. 2 20. csc y + cot y = Vs. y = 2mr + -■ o 21. 3 (sec 2 a + cot 2 a) = 13. a = nir ± -, nir ± - ■ o o Find all the positive angles less than 360° which satisfy the following equations : 22. cos 2x + cosx = - 1. Ans. x = 90°, 120°, 240°, 270°. 23. sin 2 x — cos 2 x — sin x + cos x = 0. x = 0°, 90°, 210°, 330°. V3 24. sin (60° - x) - sin (60° + x) = x = 240°, 300°. A V3 25. sin (30° + x) - cos (60°+ x) = x = 210°, 330°. 26. tan (45° - x) + cot (45° - x) = 4. x = 30°, 150°, 210°, 330°. 27. cos 2 x = cos 2 x. x = 0°, 180°. 28. 2 sin y = sin 2y. y = 0°, 180°. 29. sin x + sin 2 x + sin 3 x = 0. x = 0°, 90°, 120°, 180°, 240°, 270°. 30. tan x + tan 2 x = tan 3 x. x = 0°, 60°, 120°, 180°, 240°, 300° 31. secx — cotx = cscx — tanx. x = 45°, 225°. 32. sin 4 x - cos 3 x = sin 2 x. x = 30°, 90°, 150°, 2 10°, 270°, 330°. 33. Vl + sin x — Vl — sin x = 2 cos x. 34. sin 4 x + cos 4 x = - ■ 8 35. sec(x+120°)+sec(x-120°) = 2cosx. 36. sin (x + 120°) + sin (x + 60°) = - ■ 37. sin y + sin 3 y = cos y — cos 3 y. Find the general value of x that satisfies the following equations : 38. cosx = and tanx = 1. Ans. x = (2n + lW + -■ V2 v ' 4 39. cot x = — V3 and csc x = — 2. x = 2mr — -■ 6 40. Find positive values of A and B which satisfy the equations cos(A-S) = - and sin(^+B) = l. Ans. —and-. 2 '2 12 4 41. Find positive values of A and B which satisfy the equations tan(4-B) = l and sec(^ + £) = -^- ^ns. ^1 and — . V5 24 24 CHAPTER VI GRAPHICAL REPRESENTATION OF TRIGONOMETRIC FUNCTIONS 49. Variables. A variable is a quantity to which an unlimited number of values can be assigned. Variables are usually denoted by the later letters of the alphabet, as x, y, z. 50. Constants. A quantity whose value remains unchanged is called a constant. Numerical or absolute constants retain the same values in all problems, as 2, 5, v7, ir, etc. Arbitrary constants are constants whose values are fixed in any particular problem. These are usually denoted by the earlier letters of the alphabet, as a, b, c, etc. 51. Functions. A function of a variable is a magnitude whose value depends on the value of the variable. Nearly all scientific problems deal with quantities and relations of this sort, and in the experiences of everyday life we are continually meeting conditions illustrating the dependence of one quantity on another. Thus, the weight a man is able to lift depends on his strength, other things being equal. Hence we may consider the weight lifted as a function of the strength of the man. Similarly, the distance a boy can run may be considered as a function of the time. The area of a square is a function of the length of a side, and the volume of a sphere is a function of its diameter. Similarly, the trinomial x* -7x -6 is a function of x because its value will depend on the value we assume for x, and sin.A, cos 2 A, tan — Z are functions of A. 52. Graphs of functions. The relation between the assumed values of a variable, and the corresponding values of a function depending on that variable, are very clearly shown by a geometrical representa- tion where the assumed values of the variable are taken as the abscissas, and the corresponding values of the function as the 93 94 PLANE TRIGONOMETRY ordinates of points in a plane (see § 13, p. 26). A smooth curve drawn through these points in order is called the graph of the function. Following are General directions for plotting the graph of a function. First step. Place y equal to the function. Second step. Assume different values for the variable (= x) and calculate the corresponding values of the function (= y), writing down the results in tabulated form. Third step. Plot the points having the values of x as abscissas and the corresponding values of y as ordinates. Fourth step. A smooth curve draion through these points in order is called the graph of the function. Ex. 1. Plot the graph of 2 re- 6. Solution. First step. Let y — 2 x — 6. Second step. Assume different values for x and compute the corresponding values of y. Thus, if x = 0, y = -6; x = l, y = -4; x = 2, y = -2; x = - 1, y = -8; x = -2, y = -10; etc. Arranging these results in tabulated form, the first two columns give the corresponding values of x and y when we assume positive values of x, and the J I ' X ^n — ■/• * ~ -- . — /-!■ _ T" " 1 " 1— yfl . — — ( X V X y -6 - 6 1 -4 -1 - 8 2 -2 -2 -10 3 -3 -12 4 2 -4 -14 5 4 -5 -16 6 6 -6 -18 etc. etc. etc. etc. last two columns when we assume nega- tive values of x. For the sake of sym- metry x = is placed in both pairs of columns. Third step. Plot the points found. Fourth step. Drawing a smooth curve through these points gives the graph of the function, which in this case is a straight line. GEAPHICAL EEPEESENTATION 95 Ex. 2. Plot the graph of x 2 - 2 x - 3. Solution. Firststep. Letj/=x 2 — 2x— 3. Second step. Computing y by assum- ing values of x, we find the following table of values. X V X y -3 -3 1 -4 - 1 2 -3 -2 5 3 -3 12 4 5 -4 21 5 12 etc. etc. 6 21 etc. etc. Third step. Plot the points found. Fourth step. Drawing a smooth curve -through these points gives the graph of "the function. 53. Graphs of the trigonometric functions trigonometric function we assume values for the angle ; the circular measures of these angles are taken as the ab- scissas, and the corre- sponding values of the function found from the table on p. 9 are taken as the ordinates ■of points on the graph. Ex. 1. Plot the graph •of sinx. Solution. First step. Let y = sin x. Second step. Assuming values of x differing by 30°, we calculate the corre- sponding values of y from the table on p. 9. In tabu- lating the results it will be noticed that the angles are expressed both in degree measure and in circular \YA i i -- - 1 1 1 \ f / - / / v' JL i° /! / j 1 Y' To find the graph of a X y X y 0° 0° 30° 7T 6 .50 - 30° IT 6 - .50 60° 7T 3 .86 -.- 60° 3 - .86 90° 7T 2 1.00 - 90° IT ~ 2 -1.00 120° 2tt T .86 -120° 2tt _ ~3~ - .86 150° hit IT .50 - 150° 5tj- ~6 - .50 180° 7T -180° — 7T 210° 6 - .50 -210° 7tt 6 .50 240° 47T T - .86 -240° 4?r .86 270° 3ir 2 -1.00 -270° Sir 2 1.00 300° hit IT - .86 -300° 5tt ~~3~ .86 330° Htt 6 - .50 -330° Htt 6 .50 360° 2tt -360° -2tt 96 PLANE TRIGONOMETRY measure. It is most convenient to use the degree measure of an angle when looking up its function, while in plotting it is necessary to use its circular measure. Third step. In plotting the points we must use the circular measure of the angles for abscissas. The most convenient way of doing this is to lay off distances ir = 3. 1416 to the right and left of the origin and then divide each of these into six equal parts. Then when Also when X = o, V = 0; X = 6' y = .50=AB; X = 8* y = .86 =CD; X = 7T — ) 2 y = 1.00 -EF; etc. X = - 7T ~ — ) y = - - .50= OB; etc *-x Fourth step. Drawing a smooth curve through these points, we get the graph of sin x for values of x between — 2 ir and 2 it. It is called the sine curve or sinusoid. Discussion, (a) Since sin (x ± 2 ir) = sin x, it follows that 'y = sin x = sin (x ± 2 ir), that is, the graph is unchanged if we replace x by x ± 2 w. This means, how- ever, that every point' is moved a distance 2ir to the right or left. Hence the arc PNMLO may be moved parallel to XX' until P falls at 0, N at F, M at I, etc., that is, into the position OFIJK, and it will be a part of the curve in its new position. In the case of the sine curve it is then only necessary to plot points, say, from x — — ir to x = ir, giving the arc or double undulation MLOFI. The sine curve consists of an indefinite number of such arcs extending to the right and left. (b) From the graph we see that the maximum value of sinx(=j/) is 1(=EF=QN, etc.) and the minimum value is — l(=SJ = IiL, etc.), while x can take on any value whatever. (c) Since the graph crosses the axis of x an infinite number of times, we see that the equation sin x = has an infinite number of real roots, namely, x = 0, ± 2 ir, ± 4 tt, etc. GRAPHICAL REPRESENTATION 97 54. Periodicity of the trigonometric functions. From the graph of sin x in the above example we saw that as the angle increased from to 2 7r radians, the sine first increased from to 1, then decreased from 1 to — 1, and finally increased from — 1 to 0. As the angle increased f rqin 2 it radians to 4 ir radians, the sine again went through the same series of changes, and so on. Thus the sine goes through all its changes while the angle changes 2 ir radians in value. This, is expressed by saying that the period of the sine is 2 ir. Similarly, the cosine, secant, or cosecant passes through all its. changes while the angle changes 2 it radians. The tangent or cotangent, however, passes through all its changes while the angle changes by -it radians. Hence, the period of the sine, cosine, secant, or cosecant is 2 it radians; while the period of the tangent or cotangent is it radians. As each trigonometric function again and again passes through the same series of values as the angle increases or decreases uni- formly, we call them periodic functions. 55. Graphs of trigonometric functions plotted by means of the unit circle. The following example will illustrate how we may plot the graph of a trigonometric function without using any table of numerical values of the function for different angles such as given on p. '9. Ex. 1. Plot the graph of sin x. Solution. Let y = sin x. Draw a unit circle. Divide the circumference of the circle into any number of equal parts (12 in 1. i j '' J i T> 1 i T-' / j ' ■ L -A — a i I a ■ ■ £ - *A n N 1 this case). At the several points of division drop perpendiculars to the horizon- tal diameter. Then the sine of the angle A OB, or, what amounts to the same thing, sine of arc AB = QB, sine of arc AE = NE, sine of arc A J = OJ, etc. It is evident that if we take the lengths of the arcs as the abscissas and the corresponding lengths of the perpendiculars as the ordinates of points in a plane, these points will lie on the graph of sin x. If we choose the same scale as in 98 PLANE TEIGONOMETEY Ex. 1, p. 96, the two graphs could be made to coincide, but in this example the unit of length chosen is larger. The main features of the two graphs of sin x are the same, however, the discussion being the same for both. In In Circle Graph "When x = arc zero = zero, x = arc.A.B = OA, x = axcAC = OC, x = arc^.D = OE, x = arc^4.E = OG, x = a,rc AF = 01, x = a,vc AG = OK, x = arc-4.iT = OL, In In Circle Graph y = zero = zero ; y = QB =AB; y = PC =CD; y=OD = EF; y = NE = GH; y = MF=IJ; y = zero = zero ; y = MH=LM, etc. EXAMPLES 1. Plot the graphs of the following functions : (a) x + 2. "■ ' x + 1 (o) x 2 -4x + 3. (b) 3x-6. (p) x 3 — 4 x. (e) 2x + l. (i) 2-. (d) as*. (e) x 8 . <*>x- (q) x 3 -2x+l. (j) log 10 x. (k) 2 x 2 - i. (r) x 8 -7x + 6. (1) 8-x 2 . (s) x 8 - 5x- 12 < g) x-2- (m) 6 + 5 x + x 2 . (t) x 4 -l. (n) x 2 - 3 x + 2. (u) x 5 - 2. 2. Plot the graph of cos x. Solution. Let y = cos x. The cosine curve is found to be as follows : ■jP - - tp: :: ilil | I 1 1 | ) 1 j — L -[~- > jid ]/-L i ^v t il T^ V'.^.J.,..— __.,/...,!,, , ,. ,,\. . JL- 3_^t5i.-»T • J M£s st j_ii_ -Ji^fO- - -jJ?< J L I. J\ - 4\ s SL'-t-- S * _ J? V *^^ &^ ~~[ I 1 "1 \ r ' r ■ ■ ■ ■ ! -£\\ 1 To plot the graph of cos x by means of the unit circle we may use the circle on p. 97. Taking the abscissas as arcs zero, AB, AC, AD, etc., and the corresponding ordinates as OA, OQ, OP, zero, etc., respectively, we will get points lying on the cosine curve. 3. Plot the graph of tan x. Solution. The tangent curve is shown on next page. GRAPHICAL REPRESENTATION 99 i 1 i i J 1 1 1 3 r A. . - 1 i |] I ' -jM hr tlr _ '_ 'l ' 1 ' '' j t -- i 11 : - : /- 1 I i • l 1 _/ i i /■" t" ■Is * ! i 1 gar i i r i J i i jiQl M A S=5 Hit ! -MF-f- / j.jr|_. ' L£ L w — *3«L-2,. ir P I r U :_l /"- fr- V ^_CD^ 3 X -' ! -r-i- 7h!-r 1 i i _j_ 1 li 1 : ! " J * T /-a i i / li i 1 J i / fl } ,-i/ J III ' J £ r . ,| 7i i 1 / f 1 | 7 | - J UL --- - i 1 -L L i I 1 ! ltF" -h4- i 1 _J_ i it .J — II jj 1 1 i ;| 1 4r 1 1 ji I L 1 . J_ "i J f J "T )| 1 ' ■ ■ r 1 l 1 To construct the tangent curve from the unit circle shown, we have In In In Is Circle Graph Circle Graph When i = arc zero = zero, x = a,TcAB = OA, x = arc .AC = OC, x = arc AD = OD, x = arc AE = OE, y = zero = zero ; y = AM = AB ; y = AN = CD ■ y = qo = oo ; y = AQ =EF, etc. 4. Plot the graph of sec x. Solution. The secant curve is given below. — +H ± rl "+t +m- ±rt£ +Hi^- u 1 1 1 ■ i i 1 : i 1 | II _it : 1 1 - ' - ' 4 -i - ii 1 ! ' / E i r f 1 I I ! | 1 [ 4. J r i / i 1 1 L ft v 1 l I /' H 1 ' ! 2 i 31 s I 1 V^ Va Y 4^ , 13 I" C * E" iSij -I 1j ! .4 isi_a v I"? ■*■ sT . : j r i ; 9 I' 3^ p — T- -A- ! ■ IE. _ 1 V !_■-'- . -;J-J -in - — \ i M - ! — i L'J- , \ \ % 1 t- — — J~L < Ai,-- .._ .j ! \ 1 ii ■■ljh-j-f- -f- -h-i fi-,+r iliV +-]]- ' _U i ! -'JL j 1 i . II nf y* ! 1 ! !j Using the unit circle, we have In In Circle Graph When x — arc zero = zero, x = a,rc AB = OB, x = arc-AC = OD, x = axe AD = OF, x = arcAE = OG, In In Circle Graph y = OA = OA ; y= OM=BC; y= ON -DE; y = co = oo; y = OQ = GS, etc. 100 PLANE TRIGONOMETRY 5. Plot the cotangent curve. 6. Plot the cosecant curve. - --|l-ll -'--}— - ^h--|iH-K^^^T 11 il 1 i ! L_. JiJ LL -[-- 1 ! 1 1 ! ! 1 -H"9 : -1 __IL \ . 1 IV i ! ' ij ji i ! ' ! i t I /! ill 1 - ' '-/i W-M i i 1 ! / »l 1 I. i , 1 ■ \R 1 i _fcl . PZlJ -r- -J/ 1!- U U -H-ff I ■ / H 1 • I ' ■ 1 ■ 5 B ^/ 1 i i n J-I-U-* * 1 1 » / [i .u : _ ,. :.,. !\; :_,jll|Ijjl u..: H D K >/i [' ' ! I ! : I [^a ^-^3^\ r ia: i ,l S-~ ul'-' '- I ! 'El J ?w to- u-^ar : f, ; ; IS"! ! 1 T 'V 1 i - -> 3 4. ^ II flT [ L -i w - -H - ^^ ■ ' _(. - X i 1 A i i ! U * a 3 *■<*-* I zt 1 1 , r ! ^ ]| J !' i i^**^^fc- T ^^^ I # \ i > j- [ ij 1/ i ! f I \ !■- i|_ i / \ t Ii / \ r ui-L * L 1 ii • / r ^m-hT 4 '1 i i ! L : -!-!-!- -jt-h-i H-l-hvl-l-H P-M-hlifH- 1 - 1 - 7. Draw graphs of (a) sin a; + cos s, (b) cos x — sin x, (c) sin 2 a;, (d) tan 2 a;, (e) sin x cos x. CHAPTER VII SOLUTION OF OBLIQUE TRIANGLES 56. Relations between the sides and angles of a triangle. One of the principal uses of Trigonometry lies in its application to the solution of triangles. That is, having given three elements of a triangle (sio}es and angles) at least one of which must be a side, to find the others. In Plane Geometry the student has already been taught how to solve triangles graphically. That is, it has been shown how to construct a triangle, having given Case I. Two angles and one side. Case II. Two sides and an opposite angle. Case III. Two sides and the included angle. Case IV. Three sides From such a construction of the required triangle the parts not given may be found by actual measurement with a graduated ruler and a protractor. On account of the limitations of the observer and the imperfections of the instruments used, however, the results from such measurements will, in general, be only more or less rough approximations. After having constructed the triangle from the given parts by geometric methods, it will be seen that Trigonometry teaches us how to find the unknown parts of the triangle to any degree of accuracy desired, and the two methods may then serve as checks on each other. The student should always bear in mind, when solving triangles, the two following geometrical properties which are common to all triangles : (70) The sum of the three angles equals 180°. (71) The greater side lies opposite the greater angle, and conversely. The trigonometric solution of oblique triangles depends upon the application of three laws, — the law of sines, the law of cosines, and the law of tangents, to the derivation of which we now turn our attention. 101 102 PLANE TRIGONOMETRY 57. Law of sines. The sides of a triangle are proportional to the sines of the opposite angles. Proof. Fig. 1 represents a triangle all of whose angles are acute, while Fig. 2 represents a triangle, one angle of which is obtuse (as A). Fig. 1 Fig. 2 Draw the perpendicular CD(= h) on AB or AB produced. From either figure, using the right triangle ACD, W ■ A k sm.A = -• b Tin Fig. 2, 8in^=sin(180°-^)=sinC^l-D--.l Also, using the right triangle BCD, (-B) Dividing (A) by (B) gives or, by alternation in proportion (C) Similarly, by drawing ] (D) (E) Writing (C), (D), (E) as a single statement, we get the law of sines. sin .8 = h a sin A sinB a V , ion, a b sini? sin A endiculi irs from A and B w< * c and sin.B sin C c a respectively. sinC sin A ' (72) sin A sin B sin C SOLUTION OF OBLIQUE TEIANGLES 103 Each of these equal ratios has a simple geometrical meaning, as may be shown if the Iujjo of sines is proved as follows : Circumscribe a circle about the triangle ABC as shown in the figure, and draw the radii OB, OC. Denote the radius of the circle by B. Draw OM per- pendicular to BG. Since the inscribed angle A is measured by one half of the arc BG and the central angle BOC is measured by the whole arc BC, it follows that the angle BOC = 2 A, or, angle BOM = A. Then BM = R sinBOM=R sin A, by (7), p. 11 and a = 2BM = 2RsinA, a or, 2R: sin A In like manner it may be shown that 2R = and 2K = — — . sin B sin C Hence, by equating the results, we get a be sinB 2R = sin A sinC The ratio of any side of a triangle to the sine of the opposite angle is numer- ically equal to the diameter of the circumscribed circle. It is evident that a triangle may be solved by the aid of the law of sines if two of the three known elements are a side and its opposite angle. The case of two angles and the included side being given, may also be brought under this head, since by (70), p. 101, we may find the third angle which lies opposite the given side. Ex. 1. Given A = 65°, B = 40°, a = 60 ft. ; solve the triangle. Solution. Construct the triangle. Since two angles are given we get the third angle at once from (70), p. 101. Thus, C = 180° - (A + B) = 1S0° - 105° = 75°. Since we know the side a and its opposite angle A we may use the law of sines, but we must be careful to choose such ratios in (72) that only one un- known quantity is involved. Thus, to find the side 6 use a b sin .4 sini? Clearing of fractions and solving for the only unknown quantity 6, we get t a sini? o= ■ ■ • sin A Substituting the numerical values of sin A and sin B from the table on p. 0, and a = 50 ft, we get 5 x 0.6428 6 = 0.9063 = 35.46 ft. 104 PLANE TEIGONOMETEY Similarly, to find the side c, use a c sin A sin G Clearing of fractions and solving for c, we get asinO 50 x 0.9659 sin .4 0.9063 = 53.29 ft. By measurements on the figure we now check the results to see that there are no large errors. Since we now know all the sides and angles of the triangle, the triangle is said to be solved. 58. The ambiguous case. When two sides and an angle opposite one of them are given, the solution of the triangle depends on the law of sines. We must first find the unknown angle which lies opposite one of the given sides. But when an angle is determined by its sine, it admits of two values which are supplements of each other ; hence either value of that angle may be taken unless one is excluded by the conditions of the problem. Let a and b be the given sides and A (opposite the side a) the given angle. If a>b, then by Geometry A>B, and B must be acute whatever be the value of A, for a triangle can have only one obtuse angle. Hence there is one, and only one, triangle that will satisfy the given conditions. If a = b, then by Geometry A=B, both A and B must be acute, and the required triangle is isosceles. If a<b, then by Geometry A< B, and A must be acute in order that the triangle shall be possible ; and when A is acute it is evident from the figure that the two triangles ACB and ACB' will satisfy the given conditions provided a is greater than the perpendicular CP ; that is, provided a >b smA. The angles ABC and AB'C are supplementary (since Z B'BC = Z BB'C) ; they are, in fact, the supplementary angles obtained (using the law of sines) from the formula sinB = b sin A . SOLUTION OF OBLIQUE TEIANGLES 105 That is, we get the corresponding acute value B from a table of- sines, and the supplementary obtuse value as follows : B' = 1W°-B. If, however, a = b sin A = CP, then sinB = 1, B = 90°, and the tri- angle required is a right triangle. If <z< 5 sin 4 (that is, greater than CP), then sinS>l, and the triangle is impossible. These results may be stated in compact form as follows : Two solutions : If A is acute and the value of a lies between b and b sin A. No solution : If A is acute and a <Lb sin A, or if A is obtuse and a < b or a = b. One solution : In all other cases. The number of solutions can usually be determined by inspection on constructing the triangle. In case of doubt find the value of b sin A and test as above. Ex. 1. Given a = 21, 6 = 32, A = 115° ; find the remaining parts. Solution. In this case a <b and A > 90° ; hence the triangle is impossible and there is no solution. Ex. 2. Given a = 32, 6 = 86, A = 30° ; find the remaining parts. Solution. Here 6 sin 4 = 86 x \ = 43; hence a<bsinA, and there is no solution. Ex. 3. Given a = 40, 6 = 30, A = 75° ; find the remaining parts. Solution. Since a > 6 and A is acute there is one solution only. r>y me law a b C */ sin. A sins' or, . „ 6sin^. 30 x .9659 sin B = = — a 40 .-. sinB = .7244, X or, B = 46.4°, the only admissible value of B. A\75° ??v A C; = i Then C = 180° - (A + B) = 180° - 121.4° = 58.6°. To find C, we get, by the law of sines, sin C sin A _ a sin C __ 40 x .8535 C ~~ sin^i .9659 Check the results by measurements on the figure. = 35.3. 106 PLANE TRIGONOMETRY Ex. 4. Solve the triangle, having given b = 16, a = 12, A = 52°. Solution. Here bsinA = 15 x .7880 = 11.82 ; hence, since A is acute and a lies between 6 and 6 sin .4, there are two solutions. That is, there are two tri- angles, ACBi and A CB 2 , which satisfy the given conditions. By the law of sines, a b sin A sin Bi ■ sinBi b sin A 16 x .7880 = .9850. a 12 This gives Bi = 80.07°, and the supplementary angle B 2 = 180° — B x = ! Let us first solve completely the triangle AB^C. d = 180° - (A + B{) = 47.93°. a c\ Bythe law of sines, or, sin A sinCi a sin Ci 12 x .7423 .7880 sin A Now, solving the triangle AB^C, C 2 = 180° - {A + B 2 ) = 28.07°. By the law of sines, Slll^l Bill U 2 12 x .4706 :11.3. sin A or, c 2 = The solutions then are : sinC 2 a sin C 2 sin .4. .7880 = 7.2. Tor triangle ABiC B,_= 80.07°, Ci = 47.93°, c 1 = 11.3. Tor triangle AB^C JB 2 = 99.93°, C 2 = 28.07°, c 2 = 7.2. Check the results by measurements on the figure. In the ambiguous case care should be taken to properly combine the calculated sides and angles. EXAMPLES 1. Find the number of solutions in the following triangles, having given : (a) a = 80, b = 100, A = 30°. Ans. Two. (b) o = 50, 6 = 100, A = 30°. One. (c) a = 40, 6 = 100, A = 30°. None. (d) a = 13, b = 11, A = 69°. One. (e) a = 70, 6 = 75, A = 60°. Two. (f) a = 134, 6 = 84, B = 52°. None. (g) a = 200, b = 100, A = 30°. One. 2. Solve the triangle, having given a = 50, A = 65°, B = 40°. 4ns. C = 75°, 6 = 35.46, c = 53.29. 3. Solve the triangle, having given 6 = 7.07, A = 30°, C = 105°. Ans. B = 45°, a = 5, c = 9.66. SOLUTION OF OBLIQUE TRIANGLES 107 4. Solve the triangle, having given c = 9.56, 4 = 45°, B = 60°. 4ns. C = 75°, a = 7, 6 = 8.57. 5. Solve the triangle when c = 60, A = 50°, B = 75°. Ans. = 55°, 6 = 70.7, a = 56.1. 6. Solve the triangle when a = 550, A = 10° 12', B = 46° 36'. 4ns. C = 123° 12', 6 = 2257.4, c = 2600.2. 7. Solve the triangle when a = 18, 6 = 20, A = 55.4°. 4ns. JBi = 66.2°, d = 58.4°, Cj = 18.6 ; .B 2 = 113.8°, C 2 = 10.8°, c 2 = 4.1. 8. Solve the triangle when a = 3 V2, 6 = 2 V5, 4 = 60°. Ans. C = 75°, B = 45°, c = 4.73. 9. Solve the triangle when 5 = 19, c = 18, C = 15° 49'. Ans. B x = 16° 43', A 1 = 147° 28', a t = 35.5 ; B 2 = 163° 17', 4 2 = 54', a 2 = 1.04. 10. Solve the triangle when a = 119, 6 = 97, A = 50°. 4ns. B = 38.6°, C = 91.4°, c = 155.3. 11. Solve the triangle when a = 120, 6 = 80, A = 60°. 4ns. B = 35.3°, C = 84.7°, c = 137.9. 12. It is required to find the horizontal distance from a point 4 to an inaccessible point B on the opposite bank of a river. We measure off any convenient horizontal distance as AC, and then measure the angles CAB and ACB. Let 4C=283feet, angle C45=38°, and angle 4 CB= 66. 3°. Solve the triangle ABC for the side AB. Ans. 267 .4 ft. 13. A railroad embankment stands on a horizontal plane and it is required to find the distance from a point 4 in the plane to the top B of the embankment. Select a point C at the foot of the embankment lying in the same vertical plane as 4 and B, and meas- ure the distances 4 C and CB, and the angle BAG. Let 4C = 48.5 ft, BC = 84 ft., and angle B4C = 21.5°. Solve the triangle for the side AB. Ans. 127.2 ft. 14. A tree 4 is observed from two points B and C, 270 ft. apart, on a straight road. The angle BCA is 55° and the angle CBA=6b°. Find the distance from the tree to the nearer point B. Ans. 255.4 ft. 15. To determine the distance of a hostile fort 4 from a place B, a line BC and the angles ABC and BCA were measured and found to be 1006.62 yd., 44°, and 70° respectively. Find the distance AB. Ans. 1035.5 yd. 16. A triangular lot has two sides of lengths 140.5 ft. and 170.6 ft., and the angle opposite the former is 40°. Find the length of a fence around it. 4ns. 353.9 ft., or 529.6 ft. 17. Two buoys are 64.2 yd. apart, and a boat is 74.1 yd. from the nearer buoy. The angle between the lines from the buoys to the boat is 27.3°. How far is the boat from the further buoy ? 4ns. 120.3 yd. 108 PLANE TEIGONOMETE.Y 18. Prove the following for any triangle : (a) a = b cos C + c eosB, b = a cosC + c cos A, c = a cosB + b cos A. 1 sin C + c 2 sin B (b) V&esin.BsinC7 : b + c (c) (d) sin A + 2 sin B _ sin C a + 2b c sin? A — m sin 2 B sin 2 C a 2 - m& 2 c 2 (e) a sin (B - C) + b sin (C - 4) + c sin (.4 - B) = 0. 19. If R is the radius of the circumscribed circle, prove the following for any triangle [s = \ (a + b + c)] : (a) B(smA + sinB + sin C) = s. (b) be = 4 £ 2 (cos 4 + cos B f-os C). 1 1 (o) ■ + • 1 4R s — a s — 6 s — c 20. Show that in any triangle 8 Vs (s — a) (s - b) (s — c) a + b __ cos\(A-B) c sin ^ C 59. Law of cosines. In any triangle the square of any side is equal to the sum of the squares of the other two sides minus twice the product of these two sides into the cosine of their included angle. Proof. Suppose we want to find the side a in terms of the other two sides b and c and their included angle A. t-;s- When the angle A is acute (as in Fig. 1) we have, from Geometry, UB i = AC* + AB i -2ABxAD ) [The square of the side opposite an acute angle equals the sum"] of the squares of the other two sides minus twice the product I of one of those sides into the projection of the other upon it.J or, But Hence a 2 = b 2 +c*-2cAD. AD= b cos A . a 2 = b 2 + c 2 — 2 be cos A. (8), p. 11 SOLUTION OF OBLIQUE TKIANGLES 109 When the angle .1 is obtuse (as in Fin. 2) we have, from Geometry, _.-C _b" T _i* ^ Z ^5 v - CB* =AC* +AB* + 2AB X AD, I ine square of the side opposite an obtuse angle equals the siuu"| I of the squares of the other two skies plus twice the product I Lof one of those sides into the projection of the other upon it. J or. a s =b a +i"+2cAD. But A D = b cos DA C (8\ p. 11 = &eos^l$0 : — .1) = — b cos A. Hence in any ease (73) a 3 =& s + c 2 -2dccos.i. Similarly, we may find (74) 6 s = a* + c 3 — 2 ac cos B. ^75) c* = a s + b i -2abcosC* Observe that if A = 90°. then o os -4 = 0. and (73 > becomes o 3 = b 2 4- r 2 , which is the known relation between the sides of a right triangle where -1 is the right angle. Solving (73), (J4 1 , (75 > for the cosines of the angles, we get &* + £*_<,* (76) cosA — 2 be tf + c*- & 2ac a s +6 s - -c* (77) cosB = (78) cosC = 2Qb These formulas are useful in finding the angles of a triangle, having given its sides. Formulas ^73 >. (74), (75) may be used for finding the third side of a triangle when two sides and the included angle are given. The other angles may then be found either by the law of sines or by formulas (76). (77 (78. * Since a and A. 6 and 18, e and C stand for any - "1e of a triangle and the opposite angle, from any formula expressing a general relation between those parts another formula may be deduced by i-AtinytMjj lie letter « eyelical order. Thus, in 73 by changing a to ft, 6 to c r e to t. and A to Bwv obtain 74 ; and in 74 by .'ti.mdn; 6 toco toa,a to t. and.fi to C we get 1 75 . This is a great help in mem -■rijfr.^ s. ■me sots ot" formulas. <r , '* 110 PLANE TRIGONOMETRY Ex. 1. Having given A = 47°, 6 = 8, c = 10 ; solve the triangle. Solution. To find the side o use (73). a 2 = 6 2 + c 2 - 2 6ccos4 = 64 + 100 - 2 x 8 x 10 x .6820 = 64.88. .-. a = V54.88 = 7.408. .=io -3 To find the angles C and B use the law of sines. 1nS = »^ = 8x - 7814 = .7896. ,.5 = 52.2°. a 7.408 sin C = i^i = 10 X - 7314 = .9873. ,. C = 80.8°. a 7.408 To check our work we note the fact that A + B + C = 47°+ 52.2°+ 80.8°= 180°. Ex. 2. Having given a = 7, 6 = 3, c = 5; solve the triangle. Solution. Using formulas (76), (77), (78) in order to find the angles, we get cos^ = &2 + c2 - a2 = 32 + 62 - 72 = -l = -.5000. ...1 = 120°. 2 6c 2-3-5 2 cosB = " 2 + c 2 -6 2 = 7 2 +5 2 -3 2 = 18 = 92g6 . B = 2ac 2-7-5 14 cosC= a2 + &2 - c2 = 72 + 82 - 62 = l 1 = .7857. ,. C = 38.2°. 2 ab 2-7-3 14 C/iecfc : A + B+C = 120° + 21.8° + 38.2° = 180°. EXAMPLES 1. Having given a = 30, 6 = 54, C = 46° ; solve the triangle. Ans. 4 = 33.1°, B= 100.9°, c = 39.56. 2. Having given 4 = 60°, 6 = 8, c = 5 ; find a and the cosines of the angles B and C. Ans. 7, \, \\. 3. Having given a = 33, c = 30, B = 35.4° ; find 4 and O. Ans. 4 = 80.7°, C=63.9°. 4. Having given a = 4, 6 = 7, c = 10 ; solve the triangle. 4ns. A = 18.2°, B = 33.1°, C = 128.7°. 5. Having given a = 21, 6 = 24, c = 27 ; solve the triangle. Ans. A = 48.2°, B = 68.4°, O = 73.4°. 6. Having given a = 2, 6 = 3, c = 4; find the cosines of the angles A, B, C. Ans - h ib ~ i- 7. Having given a = 77.99, 6 = 83.39, C = 72° 15' ; solve the triangle. Ans. A = 51° 15 ', B = 66° 30', c = 95.24. 8. If two sides of a triangle are 10 and 11 and the included angle is 50°, find the third side. Ans. 8.92. SOLUTION OF OBLIQUE TKIAXGLES 111 9. The two diagonals of a parallelogram are 10 and 12 and they form an angle of 49.3° ; find the sides. Ana. 10 and 4.68. 10. In order to find the distance between two objects, A and B, separated by a pond, a station C was chosen, and the distances CA = 426 yd., CB = 322. 4 yd. , together with the angle A CB = 68. 7°, were measured. Find the distance from A to B. Ans. 430.85 yd. 11. A ladder 52 ft. long is set 20 ft. from the foot of an inclined buttress, and reaches 46 ft. up its face. Find the inclination of the face of the buttress. Ans. 95.9°. 12. Under what visual angle is an object 7 ft. long seen by an observer whose eye is 5 ft. from one end of the object and 8 ft. from the other end ? Ans. 60°. 13. Two stations, A and B, on opposite sides of a mountain, are both visible from a third station C. The distance AC = 11.5 mi., BC = 9.4 mi., and angle ACB = 59.5°. Find the distance between A and B. Ann. 10.5 mi. 14. Prove the following for any triangle : (a) a(b 2 +c 2 )cosA + 6(c 2 + a 2 ) cosB+ c(a? + V>)cosC-- „ „ 6 + c cosB + cosC (b) 3a6c. a 1 — cos.4 (c) a + 6 + c = (6 + c) cos A + (c + a) cosB +(a + b) cos C. cos A . cosB cosC a 2 + 6 2 + c 2 (d) • + - a b c 2abc (e) a 2 + ft 2 + c 2 = 2 (aft cos C + bc cos A + ca cosB). 60. Law of tangents. The sum of any two sides of a triangle is to their difference as the tangent of half the sum, of their opposite angles is to the tangent of half their difference. Proof. By the law of sines, a b sin A sinj} and by division and composition in proportion, a + b sin A + sin B w sin A — sinB But from (66), p. 74, sin A + sinB (B) sinA — sinB tan \(A +B) tan £(4 -B) 112 PLANE TEIGONOMETEY Hence equating (4) and (B), we get (79) a + b = tan K^+*) * ^ ' a — b tan §(4 — S) ' a- -t i a + c tani(4+C) Similarly, we get = *-* 1» ■" 5 a-c tan£(4-C) b + c ^ tan \ (B + C) t 6 -c - tan £(.B-C)' When two sides and the included angle are given, as a, b, C, the law of tangents may be employed in finding the two unknown angles A and B.% Since a + b, a — b, A+B (=180°— C), and there- fore also tan £(4 +B), are known, we clear (79) of fractions and solve for the unknown quantity tan \ (A — B). This gives (80) tan §(4-.B) = ll-? tan 1(4+*). ' We shall illustrate the process by means of an example. Ex. 1. Having given a = 872. 5, b = 632.7, C = 80° ; solve the triangle. Solution. a + b = 1605.2, a - 6 = 239.8, A + B = 180°- C= 100°, and i {A + B) =50°. From (79), since tan £ {A + B) = tan 50° = 1.1918, tan h(A - B) = - — -tan 1(4 + B) = =^^- x 1.1918 = .1899. 5 a + b 5V ; 1505.2 .-. £ (A - B) = 10.6°. Adding this result to £ (A + B) = 50° gives A = 60.6°. Subtracting the result from ^ (A + B) = 50° gives B B = 39.4°. To find the side c, use the law of sines. Thus, asinC 872.5 x .9848 no „ „ c = — = = yob./. sinJ. .8712 We will now derive formulas for solving triangles having three sides given, which are more convenient than (76), (77), (78), p. 109. * If b > a, then B>A, making a - & and A - B negative. The formula still holds true, but to avoid negative quantities it is better to write the formula in form -, = -' — i\-= ~ * ° b-a tan %(B -A) t These may also be found by changing the letters in cyclical order (see footnote, p. 109). % When logarithms are used in solving triangles, having given two sides and the included angle, the law of tangents, which involves products, is to be preferred to the law of cosines, which involves sums. SOLUTION OF OBLIQUE TRIANGLES 113 61. Trigonometric functions of the half angles of a triangle in terms of its sides. Denote half the sum of the sides of a triangle (i.e. half the perimeter) by s. Then (4) 2 s = a + b + c. Subtracting 2 c from both sides, 2s — 2c = a + b + c — 2c, or, (-B) 2 (s - c) = a + b - c. Similarly, (C) 2(s- b)=a-b + c, (D) 2(s — a) = —a + b + c. In (49a), (49b), p. 70, replace 2x by A, and, what amounts to the same thing, x by \A. This gives (E) 2sin 2 J^=l-cos^, (F) 2cos 2 £.4=l+eos,4. b 2 + c 2 — a 2 But from (76), p. 109, cos^l = — ! hence (E) becomes' (fl) 2ain .^ =1 _L±i!_JL _ 2Sc-ft 2 -c 2 + a il ~~ 2 be ^ a 2 -(b 2 -2bc + c 2 ) ~ 2bc _ g' _ (b - c ) 2 ~ 2 6c _ (a + b — c)(a — b + c) ~ 2 be [a2 — (5 — c) 2 being the product of the sum and difference of a and 6 — c] 2(s-c)2(s-i) , ._ ._ ( 81) .-. 8 inM = ^ME± 114 PLANE TRIGONOMETRY Similarly, (F) becomes J2 + c 2 _ a 2 2 be + b 2 + e 2 - a 2 ~ 2 be _ (b + e) 2 - a 2 2 be _ (b + e + a)(b + c - a) 2bc _ 2s-2(s-a) 2 be (82) .:cosU = ^EE^. citi JL A Since tan \A — t—-> we get, by substitution from (81) and (82), COS ft- J*. (83) tanM.xBS. v ' N s(s— a) Since any angle of a triangle must be less than 180°, \A must be less than 90° and all the functions of \A must be positive. Hence only the positive signs of the radicals in (81), (82), (83) have been taken. Similarly, we may get sin i5 = ^3S3, sin^= > J \s(s-b) f tan \B = (s — a)(s — c) |( s - a )( s " -b) i ab \s(s-e) I ab ' (s — a)(s - -b) »"-Y s(s-b) ' tan ^ C= N ,( s _ ) There is then a choice of three different formulas for finding the value of each angle. If half the angle is very near 0°, the formula for the cosine will not give a very accurate result, because the cosines of angles near 0° differ little in value ; and the same holds true of the formula for the sine when half the angle is very near 90°. Hence in the first case the formula for the sine, in the second that for the cosine, should be used. In general, however, the formula for the tan- gent is to be preferred. * Also found by changing the letters in cyclical order. SOLUTION OF OBLIQUE TEIANGLES 115 When two angles, as A and B, have been found, the third angle, C, may be found by the relation A+B + C = 180°, but it is best to compute all the angles from the formulas, so that we use the sum of the angles as a test of the accuracy of the results. It is customary to use a second form of (83), found as follows : * \ s(s — a) = J . / T - g )Q-6)0 Z "c) ' \ s(s-a) 2 [Multiplying both numerator and denomina-l tor of the fraction under the radical by s~ a. J _ l 1 -«)(s- &)0-0 s — a N s Denoting the radical part of the expression by r, (84) r = -\J-i ^ ii 1 , and we get (85) tan|A= — - — Similarly, (86) tan|B = s—b r (87) tan | C = By proving one of the last three formulas geometrically it may be shown that r is the radius of the inscribed circle. C Proof. Since angle NA = \A, NO M A (A) tan^: ' AN If « denotes half the perimeter, we have 2s=AN+NB+BL+LC+CM+MA. A ^ ^J? But NB = BL, CM = LC, MA = AN; therefore 2 s = 2 AN + 2BL + 2LC, or, s = AN+(BL + LC) = AN + a. This gives AN = s — a. NO Substituting in (A), tan J A = Comparing this result with (85) and (84) shows that N Q =r = J {S ~ a) {S ~ b) (8 ~~^ * When logarithms are used in solving triangles, having given the three sides, formulas (84). (85). (86). (87), which involve products, are more convenient than the law of cosines, which involves sums. 116 PLANE TRIGONOMETRY Ex. 1. Solve the triangle whose sides are 13, 14, 16. Solution. Let a = 13, 6 = 14, c = 16. Then 2s = a + 6 + c = 42, or, s = 21. Also, s — a = 8, s — 6 = 7, s — c = 6. From (84), r _ /(s-a)(s-6)(s-c)_ te.7.6 From (85), tani^= r = 4 = 1 =.5000. 2 o „ B <> . . J 4 = 26.56°, or A = 53.12°. From (86), tan 1 B = ~^— = - = .5714. v " * s-6 7 .-. |S= 29.74°, or B = 59.48°. From (87), tan 1 C = — ^— = - = - = .6667. v " * s-c 6 3 . . £ C = 33.69°, or C = 67.38°. Cftecfc : 4 + B + C = 53. 12° + 59.48° + 67.38° = 179.98°.* EXAMPLES 1. Solve Examples 1, 3, 8, p. 110, using the law of tangents. 2. Solve Examples 4, 5, 6, p. 110, using formulas (84), (86), (86), (87), p. 116. 3. Prove the following for any triangle : (a) (a + b) sin J. C = c cos \{A-B). (b) tan£Btan^C = 6 + C ~ a . i 3 6 + c + a (c) 6 cos 2 £ + c cos 2 ^B = s. (d) (6 + c - a) tan ^ A = (c + a - 6) tan ^ B. (e) & = (a + o) 2 sin 2 J C + (a - 6) 2 cos 2 £ C. (f) c (cos A + cosB) = 2 (a + 6) sin 2 ^ C. cos 2 £ A _ a (s - a) ™ cos 2 £ B _ 6 (s - 6) ' (h) 6 sin 2 — he sin 2 — = s — a. (i) a cos £ 2? cos J C esc ^ A = s. (j) sin^d. = 2 sin — cos — = — Vs (s — a)(s — b) (s — c). * The error .02° arises from the fact that we used a four-place table. If we had used a table giving the tirst five significant figures of the tangent, the error would have been less ; if a six-place table, still less, etc. For ordinary purposes, however, the results we get, using a four-place table, are sufficiently accurate. SOLUTION OF OBLIQUE TRIANGLES 11' 4. If B and r denote the radii of the circumscribed and inscribed circles respectively, prove the following for any triangle : o sin £ B sin £ C cos^A abe i ,f (a) r = (b) R. (c) r + — +^ = r-^- oc ca ao 2riJ a&c 4V*(s- a)(s- 6)(s -c) 2 Ysin.4 sinBsinC (e) after = 4 B (s — a) (s — 6) (s — c). 62. Formulas for finding the area of an oblique triangle. Case I. When two sides and the included angle are known. Let b, c, and A be known. Take c as the base. Denote the altitude by h and the area by S. Then, by Geometry, S = \ ch. But h = b sin A (from (7), p. 11); hence (88) Similarly, S = \ be sin A. S = \ac sin B = \ ab sin C. c a The o,rea, of a triangle equals half the product of any two sides ■multiplied by the sine of the included angle. Ex.1. Find the area of a triangle, having given 6 = 20 in. , c = 15 in. , A = 60°. Solution. Substituting in (88), 1 6c sin A = - x 20 x 15 x — = 75 Vz sq. in. Ans. 2 2 2 Case II. When the three sides are known. sin A = 2 sin J^4 cos \ A ](s-b)(s-c) ^ \s(s-a) 2 / = ^ V *( s - a )0- *) («-")■ Substituting this value of sin ^4 in (88), we get (89) s = ^s(s-a)(s- b)(s-c). Ex. 2. Having given a = 13, 6 = 14, c = 15 ; find the area. Solution. S = ^ (a + b + c) = 21, s — a = 8, s-b = 7, 8 — c = 6. Substituting in (89), 8 = V*(*-a)(8-&)(8-c) = V21 x 8 x 7 x6 = 84. Ans. (51), p. 72 i ( 81 )> ( 82 ). \ pp. 113, 114 118 PLANE TRIGONOMETRY Case III. Problems which do not fall under Cases I or II directly may be solved by Case I, if we first find an additional side or angle by the law of sines. Ex. 3. Given a = 10 Vs, b = 10, A = 120° ; find the area of the triangle. Solution. This does not come directly under either Case I or Case II, but, by the law of sines, bsinA_W x £V3 1 sin B = ■ 10V3 2 Therefore B = 30° and C = 180° - (A + B) = 30°. Since we now have the two sides a and 6 and the included angle C, the prob- lem comes under Case I, and we get S = ^a6sinC= l x 10V3X 10 x ^ = 25V3. Ans. EXAMPLES 1. Find the areas of the following triangles, having given Ans. 240. 17.32. 193.18. 30 V3. 600. 17*. 15,541.7. 6V6. 30,600. 11,981 or 2347.8. 45.75. 10.4. 2. Show that the area of a parallelogram equals the product of any two adja- cent sides multiplied by the sine of the included angle. 3. Find a formula for the area of an isosceles trapezoid in terms of the par- allel sides and an acute angle. 4. Show that the area of a quadrilateral equals one half the product of its diagonals into the sine of their included angle. 5. The base of an isosceles triangle is 20, and its area is 100 -=- V3; find its angles. 6. Prove the following for any triangle : , . „ abc 2 abc I A B C\ (a) S = -— . (e) S = I cos — cos — cos — 1 • v 4fl w a + 6 + c\ 2 2 2/ (b) S = rs. a s 62 (c) S = Rr( S inA + sin B+ sin C). <*> S = 1 sin 2 B + J sin 2A ' (d) S = £ a 2 sin B sin C csc 4. (a) a = 40, 6 = 13, c = 37. (b) 6 = 8, c = 5, A = 60°. (c) 6 = 10, c = 40, A = 75°. (d) a = 10, 6 = 12, C = 60°. (e) a = 40, c = 60, B = 30°. (f) a = 7, c = 5V2, B = 135°. (g) 6 = 149, A = 70° 42', , B = 39° 18' (h) a = 5, 6 = 6, e = 7. (i) a = 409, 6 = 169, c = 510. (j) a = 140.5, 6 = 170.6, A = 40°. (k) c = 8, J? = 100.1°, C = 31.1°. (1) a = 7, c = 3, A = 60°. CHAPTER VIII THEORY AND USE OF LOGARITHMS 63. Need of logarithms* in Trigonometry. Many of the problems arising in Trigonometry involve computations of considerable length. Since the labor connected with extensive and complicated calcula- tions may be greatly lessened by the use of logarithms, it is advan- tageous for us to use them in much of our trigonometric work. Especially is this true of the calculations connected with the solution of triangles. We shall now give the fundamental principles of log- arithms and explain the use of logarithmic tables. Definition of a logarithm. The power to which a given number called the base must be raised to equal a second number is called the logarithm of the second number. Thus, if (A ) b x = N, (exponential form) then x = the logarithm of N to the base b. This statement is written in abbreviated form as follows : (B) x = logi,iV. (logarithmic form) (A) and (B) are then simply two different ways of expressing the same relation between b, x, and N. (A) is called the exponential form. (B) is called the logarithmic form. The fact that a logarithm is an exponent may be emphasized by writing (^1) in the form (base) Iog = number. For example, the following relations in exponential form, namely, 3 2 =9, 2 5 = 32, a) 8 =i, *"=», are written respectively in the logarithmic form 2 = log 3 9, 5 = log 2 32, 3 = logjl, y=log x z; * Logarithms were invented by John Napier (1550-1617), Baron of Merehiston in Soot- land, and described by him in 1614. 119 120 PLANE TRIGONOMETRY where 2, 5, 3, y are the logarithms (exponents), 3, 2, £, x are the bases, and 9, 32, \, z are the numbers respectively. Similarly, the relations 25* = V25 = 5, io-s^ = _*_ = . 001, 8' = -V& = S/64 = 4, 6°=^ = 1 are written in logarithmic form as follows : ^=log 25 5, -3 = log 10 .001, | = log 8 4, = log b L EXAMPLES 1. In the following name the logarithm (exponent), the base, and the number, and write each in logarithmic form : 2 3 = 8, i 2 = 16, 5 2 = 25, 3 8 = 27, 3* = 81. Solution. In the first one, 3 = logarithm, 2 = base, 8 = number ; hence log 2 8 = 3. Ana. 2. Express the following equations in logarithmic form : (y) 2 = ^, "v / 125 = 5, 2-* = -^, 10-* = .01,p»=g. 3. Express the following equations in the exponential form : log 4 64 = 3, log 7 49 = 2, log 6 216 = 3, log™ .0001 = - 4, log 4 2 = J, log„a = l, log o l = 0, logi,a = c. 4. When the base is 2, what are the logarithms of the numbers 1, 2, |-, 4, ^, 8, 64, 128 ? 5. When the base is 5, what are the logarithms of the numbers 1, 5, 25, 125, 6. When the base is 10, what are the logarithms of the numbers 1, 10, 100, 1000, 10,000, .1, .01, .001, .0001? 7. When the base is 4 and the logarithms are 0, 1, 2, 3, — 1, — 2, £, what are the numbers ? 8. What must be the bases when the following equations are true : Jog64 = 2? logl21 = 2? log625 = 4? log3jL = _2? 9. When the base is 10, between what integers do the logarithms of the fol- lowing numbers lie : 83, 251, 1793 ? Solution. Since logi 10 = 1 and log 10 100 = 2, and 83 is a number lying be- tween 10 and 100, it follows that logi 83 = a number lying between 1 and 2. Similarly, logi 251 = a number lying between 2 and 3, logiol793 = a number lying between 3 and 4. THEORY AND USE OF LOGARITHMS 121 10. Verify the following : (a) log 10 1000 + logiolOO + logiolO + log 10 l = 6. (*>) loglOiV + lo glOTTTT - lo glO T tfV?F = °- (c) logio.001 - log 10 .01 + log 10 .l = - 2. (d) log 2 8 - 3 log 8 2 + log 2 1 = 2. (e) 2 log a a + 2 log a - + log a 1 = 0. a (f) 2 log 4 2 + £ log 2 4 - log 2 2 = 1. (g) logs 3 + log 8 £ - logs 81 = - 5. (h) 3 log 27 3 - £ log 8 27 + log 9 3 = £. (i) 4 logi 6 4 + 2 logt-jlj + £ log 2 16 = 0. (j) 2 log 8 64 - log 7 49 + £ log 55 V = 1. (k) log 8 64 + log 4 64 + log 2 64 = 11. (1) log 5 25 - log 5 125 + 2 log 5 5 = 1. (m) 2 log 36 6 - log 6 36 + logs ■& = - 3. 64. Properties of logarithms. Since a logarithm is simply a new name for an exponent, it follows that the properties of logarithms must be found from the laws in Algebra governing exponents. Theorem I. The logarithm of the product of two factors equals the sum of the logarithms of the two factors. Proof. Let the two factors be M and N, and let x and y be their logarithms to the common base b. Then (A) log„M=x, and \og b N=y. Writing these in the exponential form, (B) b x = M, and b« = N. Multiplying together the corresponding members of equations (B), b x+ »=MN. Writing this in the logarithmic form gives log b MN = x + y = log b M + log^iV. from (A) By successive applications this theorem may evidently be ex- tended to the product of any number of factors as follows : log b MNPQ = log b M- NPQ = log, M + log b NPQ . Th. I = log 6 M+ log h N + log„PQ = log b M +- log 6 2\r+ log„P + log b Q. 122 PLANE TRIGONOMETRY Theorem II. The logarithm, of the quotient of two numbers is equal to the logarithm of the dividend minus the logarithm of the divisor. Proof. As in Theorem I, let (A) \og b M=x, and log b N=y. Writing these in the exponential form, (B) b x = M, and b« = N. Dividing the corresponding members of equations (j5), we get M N Writing this in logarithmic form gives M lo g& ^ = x - V = l°g b ikf - tog b N. from (A) Theorem III. The logarithm of the pth power of a number is equal to p times the logarithm of the number. Proof. Let log b N=x. Then b x = N. Raising both sides to the pth power, ftp* _ ]SfP_ Writing this in logarithmic form gives log;,iV p = px = p\og b N. Theorem IV. The logarithm of the rth root of a number is equal to the logarithm, of the number divided by r. Proof. Let log 6 iV=x. Then b x = N. Extracting the rth root of both sides, X 1 V = N r . Writing this in logarithmic form gives \ogJ = X - = l ^i^-log b N. r r r from the preceding four theorems it follows that if we use the logarithms of numbers instead of the numbers themselves, then the operations of multiplication, division, raising to powers, and extract- ing roots are replaced by those of addition, subtraction, multiplicar tion, and division respectively. THEORY AND USE OF LOGARITHMS 123 Ex. 1. Find the value of logio V.Q01. Solution, log™ V.001 = £ log l0 .001 Th. IV b la s (c + d)^ Ex. 2. Write logi^ — — - — - in expanded form. Solution. log 6 <fe±^ = Ilog 6 ^±^ = i \ loga a 3 + log 6 (c + d)» - log 6 c 2 j Th. I, II = J{31ogio+Jlogi(c + (i)-21og»c^. Th. Ill, IV "When no base is indicated we mean that the same base is to be used through- out. Thus, the relation Th. IV log JJ^+rt = 1 j 3 loga + l log(c + d) _ 2 logc | holds true for any number used as the base. For the sake of convenience we shall call the left-hand member of an equation like the last one the contracted form of the logarithmic expression, and the right-hand member the expanded form. Ex. 3. Write 3 log (x + 1) + 3 log (x - 1) + \ log x - 2 log (x 2 + 1) in the con- tracted form. Solution. 3 log (x + 1) + 3 log (x — 1) + \ log x - 2 log (x 2 + 1) = log(x + l) 3 + log(x - l) 3 + logx* - log(x 2 + l) 2 . (X + l) 3 (x-l) 3 X* . Vx(x 2 -1) 3 „ = log — = log '— ■ Ans. (x 2 +l) 2 s (x 2 + l) 2 Another form of the answer is found as follows : . ^fr 2 - 1 ) 3 _ ,„„ /*(s 2 -l) 6 \*_ 1 ,„„ x (*» - I) 6 (x 2 +l) 2 5 \(x 2 + l) 4 / 2 8 (X 2 + l)* EXAMPLES 1. Verify the following : (a) logm VlOOO + log 10 Vm = \. (e) log 2 V§ + log s (J) 2 = - £. (b) log 10 (. I) 4 - log 10 VM\ = - 3. (f ) log 2 (. 5)» - log 4 -^16 = - J^. (c) logio V^ + i ogl0 VlO = 0. (g) logs ^*X> + logn -2 / 121 = J„a. (d) log 10 ^T00-log 10 (.01) 2 = ^. (h) log 8 (2)s + log, (^)1=1. 2. Write the following logarithmic expressions in expanded form : * ... . aft sin C ... . Vp 2 (1 - q) (b) log (f) log (c) logP(l + r)». (g) log Vp(l + q) (m + n) 8 2 Vm — n(l + s) tVd *\ c Vo-6/ a 3 6 2 c* * To verify your results, reduce them back to the original form. 124 PLANE TRIGONOMETRY 3. Write the following logarithmic expressions in contracted form : (a) 21ogx + £log2/ - 31ogz. (b) 31og(l - x) - 2 log (2 + x) + logc. (c) glog(x - 1) - 1 logx - £ log (s + 2) + logc. (d) log y - £ log (y« + 4) + log c. (e) £{21og(s - 1) + 31og(x + 1) + £logx - f log(x2 + 1)|. 65. Common* system of logarithms. Any positive number except unity may be taken as the base, and to every particular base chosen there corresponds a set or system of logarithms. In the common system the base is 10, being the one most convenient to use with our decimal system of numbers. In what follows the base is usually omitted when writing expressions in the logarithmic form, the base 10 being always understood. Thus log 10 100=2 is written log 100 =2, etc. The logarithm of a given number in the common system is then the answer to the question : What power of 10 will equal the given number ? The following table indicates what numbers have integers for logarithms in the common system. Exponential Form Logarithmic Form Since 10 4 =10,000 we have log 10,000 = 4 10 3 =1000 log 1000 = 3 10 2 =100 log 100 = 2 10 1 =10 log 10 = 1 10° =1 logl = 10" 1 = .l log.l = -1 10- 2 = .01 log .01 = -2 io- 3 = .001 log .001 = -3 10-* = .0001 log .0001 = -4 etc., etc. Assuming that as a number increases its logarithm also increases, we see that a number between 100 and 1000 has a logarithm lying between 2 and 3. Similarly, the logarithm of a number between .1 and .01 has a logarithm lying between — 1 and — 2. In fact the logarithm of any number not an exact power of 10 consists, in gen- eral, of a whole-number part and a decimal part. * Also called the Briggs System, from Henry Briggs (1556-1631), professor at Gresham College, London, and later at Oxford. He modified the new invention of logarithms so as to make it convenient for practical use. THEORY AND USE OF LOGARITHMS 125 Thus, since 4587 is a number lying between 10 8 and 10 4 , we have log 4587 = 3 + a decimal. Similarly, since .0067 is a number lying between 10 -3 and 10 -2 , log .0067 = - (2 + a decimal) = — 2 — a decimal. For practical reasons the logarithm of a number is always written in such a form that the decimal part is positive. When the loga- rithm as a whole is negative, the decimal part may be made positive by adding plus unity to it. Then, so as not to change the value of the logarithm, we add minus unity to the whole part. Thus in the last example, , A/ .„_ , „ s , . ,, ^ ' log .0067 = (— 2) + (— a decimal) = (— 1 — 2) + (1 — a decimal) = — 3 + a new decimal. To emphasize the fact that only the whole part of a logarithm is negative, the minus sign is usually written over the whole part. For example, log .004712 = - 2.3268 = - 2 - .3268 = (_l_2) + (l-.3268) = 3.6732. The whole-number part of a logarithm is called the characteristic of the logarithm. The decimal part of a logarithm is called the mantissa of the logarithm. Thus if log 357 = 2.5527 and log .004712 = 3.6732, 2 and - 3 are the characteristics and .5527 and .6732 the mantissas. From the previous explanations and by inspection of the table on the opposite page we get the following : 66. Rules for determining the characteristic of a logarithm. The characteristic of a number greater than unity is positive, and one less than the number of digits in the number to the left of the decimal point. The characteristic of a number less than unity is negative, and is one greater numerically than the number of zeros between the decimal point and the first significant figure of the number. Ex. Write down the characteristics of the logarithms of the numhers 27,683, 466.2, 9.67, 436,000, 26, .04, .0000612, .7963, .8, .0012. Ans. 4, 2, 0, 6, 1, - 2, - 6, - 1, - 1, - 3. 126 PLANE TRIGONOMETRY Theorem V. Numbers with the same significant part * (and which therefore differ only in the position of the decimal point) have the same mantissa. Proof. Consider, for example, the numbers 54.37 and 5437. Let 10* = 54.37. If we multiply both members of this equation by 100 (= 10 2 ), we have 102 . 1Qx = 1Qx+2 = 5437; or, x + 2 = log 5437. Henee the logarithm of one number differs from that of the other merely in its whole part (characteristic). Thus, if log 47,120 = 4.6732, then log 47.12 = 1.6732, and log .004712 = 3.6732. Special care is necessary in dealing with logarithms because of the fact that the mantissa is always positive, while the character- istic may be either positive or negative. When the characteristic is negative it is best for practical reasons to add 10 to it and write — 10 after the logarithm, thus giving the logarithm a new form without change of value. Thus, if (A) log .0249 = 2.3962, we add 10 to — 2, giving 8 in the place of the characteristic, and counteract this by writing — 10 after the logarithm ; that is (B) log .0249 = 8.3962 - 10. In case we wish to divide a logarithm having a negative character- istic by an integer (as is sometimes required in applying Theorem IV, p. 122), it is convenient to add and subtract 10 times that integer. Thus in case we wish to divide such a logarithm by 2, we add and subtract 20 ; if by 3, we add and subtract 30 ; and so on. Suppose we want to divide the logarithm of .0249, which is 2.3962, by 3. We would then add and subtract 30, so that (C) log .0249 = 28.3962 - 30, a form more convenient than (A ) or (S) when we wish to divide the logarithm by 3. Thus, i log .0249 = $ (28.3962 - 30) = 9.4654 - 10. * The significant part of a number consists of those figures which remain when we ignore all initial and final zeros. Thus, the significant part of 24,000 is 24 ; of 6.050 is 605 ; of .00907 is 907 ; of .00081070 is 8107. THEORY AND USE OF LOGARITHMS 127 This result may be written in form (^4) by adding the 9 in front to the — 10 at the end, giving — 1 = 1 as the characteristic. Hence $ log .0249 = 1.4654. Another method for dividing a logarithm which has a negative characteristic will now be illustrated. Suppose we wish to divide 2.3962 (=- 2 + 0.3962) by 2. We get at once 21-2 + 0.3962 - 1 + 0.1981 = 1.1981. in case we wish to divide by 3 (as in the above example), we first add and subtract 1 in order to make the negative characteristic exactly divisible by 3. Thus, 3 |-3 + 1.3962 - 1 + 0.4654 = 1.4654. The following examples will illustrate the best methods for performing the four fundamental operations of Arithmetic on logarithms. Case I. Addition of logarithms. (a) To add two logarithms having positive characteristics, as 3.2659 and 1.9866. 3.2659 1.S 5.2525 This is in no way different from ordinary addition. (b) To add two logarithms, one having a negative characteristic, as 4.2560 and 2.8711. 4.2560 or, 6.2560 - 10 2.8711 2.8711 1.1271 9.1271-10 i.e. 1.1271 Since the mantissas (decimal parts) are always positive, the carrying figure 1 from the tenth's place is positive. Hence in adding the first way, 1— 4 + 2 = — 1=1 will be the characteristic of the sum. (c) To add two logarithms having negative characteristics, as 2.4069 and 1.9842. 2.4069 or, 8.4069 - 10 1.9842 9.9842 - 10 2.3911 18.3911 - 20 i.e. 2.3911 128 PLANE TRIGONOMETRY Case II. Subtraction of logarithms. (a) To subtract logarithms having positive characteristics. From 5.6233 From 2.4673 or, 12.4673-10 ss 3.8890 take 3.7851 3.7851 1.7343 2.6822 8.6822 - 10 i.e. 2.6822 In the first example we have ordinary subtraction. In the second we subtract a greater logarithm from a smaller one and the result as a whole is negative, (b) To subtract logarithms having negative characteristics. From take 2.1163 3.4492 4.6671 or, 12.1163-10 7.4492 - 10 4.6671 From take 1.6899 1.9083 or, 9.6899 - 10 1.9083 3.7816 i.e. 7.7816 - 10 3.7816 From take 2.1853 1.7442 or, 18.1853-20 9.7442-10 2.4411 i.e. 8.4411 - 10 2.4411 Case III. Multiplication of logarithms by numbers. Multiply by 0.6842 5 by Multiply 2.7012 3 8.7012 - 10 3 3.4210 4.1036 i.e. 26.1036-30 4.1036 In the second example the carrying figure from tenth's place is + 2. Add- ing this + 2 to — 2 x 3 gives 2 — 6 = — 4 = 4 for the characteristic. ■. Case IV. Division of logarithms by numbers. (a) Divide 3.8530 by 2. 2 1 3. 8530 1.9265 (b) Divide 2.2411 by 3. Here we first add and then we subtract 30, writing the logarithm in the form 28.2411-30. 3128.2411-30 i.e. 9.4137 1.4137 10 67. Tables of logarithms. The common system (having the base 10) of logarithms is the one used in practical computations. For the convenience of the calculator the common logarithms of numbers up to a certain number of significant figures have been computed and arranged in tabulated forms called logarithmic tables. The common system has two great advantages. THEORY AND USE OF LOGARITHMS 129 (^4) The characteristic of the logarithm of a number may be written down on mere inspection by following the rules on p. 125. Hence, as a rule, only the mantissas of the logarithms of numbers are printed in the tables. (B) The logarithms of numbers having the same significant part have the same mantissa (Th. V, p. 126). Hence a change in the position of the decimal point in a number affects the characteristic alone, and it is sufficient to tabulate the mantissas * of integers only. Thus, log 3104 = 3.4920, log 31.04 = 1.4920, log .03104 = 2.4920, log 310,400 = 5.4920 ; in fact, the mantissa of any number whatever having 3104 as its significant part will have .4920 as the mantissa of its logarithm. Table I, pp. 2, 3,f gives immediately the mantissas of the loga- rithms of all numbers -whose first significant figure is 1 and whose significant part consists of four or fewer digits ; and on pp. 4, 5 are found the mantissas of the logarithms of all numbers whose first significant figure is greater than 1 and whose significant part cpn- sists of three or fewer digits. 68. To find the logarithms of numbers from Table I, pp. 2-5. When the first significant figure of the number is 1, and there are four or fewer digits in its significant part, follow Rule I. First step. Determine the characteristic by inspection, using the rule on p. 125. Second step. Find in the vertical column N, Table I, pp. 2, S, the first three significant figures of the number. The mantissa required is in the same horizontal row with these figures and in the vertical column having the fourth significant figure at the top (and bottom). Ex. 1. Find log 1387. Solution. First step. From the rule on p. 125 we see that the characteristic will be + 3, that is, one less than the number of digits (four) to the left of the decimal point. Second step. On p. 2, Table I, we find 138 in column N. The required man- tissa will be found in the same horizontal row with 138 and in the vertical col- umn which has 7 at the top. This gives the mantissa .1421. Therefore log 1387 = 3.1421. Ans. * In order to save space the decimal point in front of each mantissa is usually omitted In the tables. t The tables referred to in this book are Granville's Fow-Place Tables of Logarithms (Ginn & Company). 130 PLANE TRIGONOMETRY If the significant part of the number consists of less than four digits, annex zeros until you do have four digits. Ex. 2. Find log 17. Solution. First step. By the rule on p. 125 the characteristic is found to be 1. Second step. To find the mantissa of 17 we look up the mantissa of 1700. On p. 3, Table I, we locate 170 in column N. The required mantissa is found in the same horizontal row with 170, and in the vertical column having at the top. This gives the mantissa .2304. Therefore log 17 = 1.2304. Ans. Ex. 3. Find log. 00152. Solution. First step. By the rule on p. 125 we find that the characteristic is — 3, that is, negative, and one greater numerically than the number of zeros (two) immediately after the decimal point. Second step. Locate 152 in column N, Table I, p. 3. In the same horizon- tal row with 152 and in the vertical column with at the top we find the required mantissa .1818. Therefore log.00152 = 3.1818 = 7.1818 - 10. Ans. To find the logarithm of a number when the first significant figure of the number is greater than 1 and there are three or fewer digits in its significant part, follow Rule II. First step. Determine the characteristic by rule on p. 125. Second step. Find in the vertical column N, Table I, pp. 4, 5, the first two significant figures of the number. The mantissa required is in the horizontal row with these figures and in the vertical column having the third significant figure at the top (and bottom). Ex. 4. Find log 5.63. Solution. First step. The characteristic here is zero. Second step. On p. 4, Table I, we locate 56 in column N. In the horizon- tal row with 56 and in the vertical column with 3 at the top we find the required mantissa .7505. Therefore log 5.63 = 0.7505. Ans. If the significant part of the number consists of less than three digits, annex zeros until you do have three digits. Ex. 5. Find log 460,000. Solution. First step. The characteristic is 5. Second step. On p. 4, Table I, we locate 46 in column N. In the horizontal row with 46 and in the vertical column with at the top we find the required mantissa .6628. Therefore log 460,000 = 5.6628. Ans. THEORY AND USE OF LOGARITHMS 131 Ex. 6. Find log. 08. Solution. First step. The characteristic is — 2. Second step. Using 800, we find that the mantissa is .9031. Therefore log .08 = 2.9031 = 8.9031 - 10. Ans. Ex. 7. Find (a) log 1872, (b) log 5, (c) log .7, (d) log 20,000, (e) log 1.808, (f) log. 000032, (g) log .01011, (h) log 9.95, (i) log 17.35, (j) log. 1289, (k) log 2500, (1) log 1.002. Ans. (a) 3.2723, (b) 0.6990, (c)_1.8461, (d) 4.3010, (e) 0.2572, (f) 5.5051, (g) 2.0048, (h) 0.9978, (i) 1.2393, (j) 1.1103, (k) 3.3979, (1) 0.0009. When the first significant figure of a number is 1 and the num- ber of digits in its significant part is greater than 4, its mantissa cannot be found in Table I ; nor can the mantissa of a number be found when its first significant figure is greater than 1 and the num- ber of digits in its significant part be greater than 3. By intei-polation,* however, we may, in the first case, find the mantissa of a number having a fifth significant figure ; and in the second case, of a number having a fourth significant figure. In this book no attempt is made to find the logarithms of numbers with more significant figures, since our four-place tables are in general accurate only to that extent. We shall now illustrate the process of interpolation by means of examples. Ex. 8. Find log 2445. Solution. By rule on p. 125 the characteristic is found to be 3. The required mantissa is not found in our table. But by Rule II, p. 130, log 2450 = 3.3892 and log 2440 = 3.3874 Difference in logarithms = .0018 Since 2445 lies between 2440 and 2450, it is clear that its logarithm must lie between 3.3874 and 3.3892. Because 2445 is just halfway between 2440 and 2450 we assume that its logarithm is halfway between the two logarithms, t We then take half (or .5) of their difference, .0018 (called the tabular difference), and add this to log 2440 = 3.3874. This gives log 2445 = 3.3874 + .5 x .0018 = 3.3883. If we had to find log 2442, we should take not half the difference, but .2 of the difference between the logarithms of 2440 and 2445, since 2442 is not half- way between them but two tenths of the way. * Illustrated by examples on pp. 16-19 In the ease of trigonometric functions. t In this process of interpolation we have assumed and used the principle that the increase of the logarithm is proportional to the increase of the number This principle is not strictly true, though for numbers whose first significant figure is greater than 1 the error is so small as not to appear in the fourth decimal place of the mantissa. For numbers whose first significant figure is 1 this error would often appear, and for this reason Table I, pp. 2, 3, gives th« mantissas of all such numbers exact to four decimal places. 132 PLANE TRIGONOMETRY In order to save work in interpolating, when looking up the loga- rithms of numbers whose mantissas are not found in the table, each tabular difference occurring in the table has been multiplied by .1, .2, .3, • • ■ , .9, and the results are printed in the large right-hand column with " Prop. Parts " (propor- tional parts) at the top. Thus, on p. 4, Table I, the first section in the Prop. Parts column shows the products obtained when multiplying the tabular differences 22 and 21* by .1, .2, .3, . - ., .9. Thus, .1 x 22 = 2.2 .1 x 21 = 2.1 .2 x 22 = 4.4 .2 x 21 = 4.2 .3 x 22 = 6.6 .3 x 21 = 6.3 .4 x 22 = 8.8 .4 x 21 = 8.4 .5 X 22 = 11.0 .5 x 21 = 10.5 etc. etc. Hence Extra Digit Difference 22 21 1 2.2 2.1 2 4.4 4.2 3 6.6 6.3 4 8.8 8.4 5 11.0 10.5 6 13.2 12.6 7 15.4 14.7 8 17.6 16.8 9 19.8 18.9 To find the logarithm of a number whose mantissa is not found in the table, t use Rule III. First step. Find the logarithm of the number, using only the first three (or four) digits of its significant part when look- ing up the mantissa.% Second step. Subtract the mantissa just found from the next greater mantissa in the table to find the corresponding tabular difference. Third step, hi the Prop. Parts column locate the block correspond- ing to the tabular difference found. Under this difference and oppo- site the extra digit § of the number will be found the proportional part of the tabular difference which should be added to the extreme right of the logarithm found in the first step. The sum will be the logarithm of the given number. Ex. 9. Find log 28.64. Solution. Since the mantissa of 2864 is not found in our table, this example comes under Rule III, the extra digit being 4. First step. log 28.60 = 1.4564 Rule II Second step. log 28.70 = 1.4579 Rule II Tabular difference = 15 || * These are really .0022 and .0021, it being customary to drop the decimal point. t That is, a number whose logarithm cannot be found by Rule I or Rule II, because its significant part contains too many digits. J When the first significant figure is 1, use the first four digits, following Rule I ; when the first significant figure is greater than 1, use the first three digits, following Rule II. § In finding log 4836, for instance, 6 is called the extra digit, or, in finding log 14,835 the extra digit is 5. II The tabular difference = .0015, but the decimal point is usually omitted in practice. THEOKY AND USE OF LOGARITHMS 133 Third step. About halfway down the Prop. Parts column on p. 4 we find the block giving the proportional parts corresponding to the tabular difference 15. Under 16 and opposite the extra digit 4 of our number we find 6.0. Then log 28.60 = 1.4564 6 Prop. Part log 28.64 = 1.4570. Ans. Ex. 10. Pind log. 12548. Solution. Since the mantissa of 12,548 is not found in our table, this example comes under Rule III, the extra digit being 8. First step. log . 12540 = 1. 0983 Second step. log . 12550 = 1.0986 Tabular difference = 3 Third step. In the Prop. Parts column on p. 2 we find the block giving the proportional parts corresponding to the tabular difference 3. Under 3 and oppo- site the extra digit 8 we find 2.4(= 2). Then log. 12540 = 1.0983 2 Prop. Part log. 12548 = 1.0985. Ans. Ex. 11. Verify the following : (a) log 4583 =3.6612. (e) log 1000. 7 =3.0003. (b) log 16.426 = 1.2155. (f) . log 724,200 =5.8598. (c) log .09688 = 2.9862. (g) log 9.496 =0.9775. (d) log. 10108 = 1.0047. (h) log .0004586 = 4.6614. 69. To find the number corresponding to a given logarithm, use Rtti/e IV. On pp. 2-5, Table I, look for the mantissa of the given logarithm. If the mantissa is found exactly in the table, the first significant figures of the corresponding number are found in the same row under the N column, while the last figure is at the top of the column in which the mantissa was found. Noting what the characteristic in the given logarithm is, place the decimal point so as to agree with the rule on p. 125. In case the mantissa of the given logarithm is not found exactly in the table we must take instead the following steps : First step. Locate the given mantissa between two mantissas in the tables. Second step. Write down the number corresponding to the lesser of the two mantissas. This will give the first three (or four) significant figures of the required number. Third step. Find the tabular difference between the two mantissas from the table, and also the difference between the lesser of the two and the given mantissa. 134 PLANE TRIGONOMETRY Fourth step. Under the Prop. Parts column find the block cor- responding to the tabular difference found. Under this tabular difference pick out the proportional part nearest the difference found between the lesser mantissa and the given mantissa, and to the left of it will be found the last {extra) figure of the number, which figure we now annex. Fifth step. Noting what the characteristic of the given logarithm is, place the decimal point so as to agree with the rule on p. 125. Ex. 12. Find the number whose logarithm is 2.1892. Solution. The problem may also be stated thus : find x, having given log x = 2. 1892. On p. 3, Table I, we find this mantissa, . 1892 exactly, in the same horizon- tal row with 154 in the N column and in the vertical column with 6 at the top. Hence the first four significant figures of the required number are 1546. Since the characteristic is 2, we place the decimal point so that there will be three digits to the left of the decimal point, that is, we place it between 4 and 6. Hence x = 154.6. Ans. Ex. 13. Find the number whose logarithm is 4.8409. Solution. That is, given log * = 4.8409, to find x. Since the mantissa .8409 is not found exactly in our table, we follow the last part of Rule IV. First step. The given mantissa, .8409, is found to lie between .8407 and .8414 on p. 4, Table I. Second step. The number corresponding to the lesser one, that is, to .8407, is 693. Third step. The tabular difference between .8407 and .8414 is 7, and the dif- ference between .8407 and the given mantissa .8409 is 2. Fourth step. In the Prop. Parts column under the block corresponding to the tabular difference 7, we find that the proportional part 2.1 is nearest to 2 in value. Immediately to the left of 2.1 we find 3, the (extra) figure to be annexed to the number 693 found in the second step. Hence the first four significant figures of the required number are 6933. Fifth step. Since the characteristic of the given logarithm is 4, we annex one zero and place the decimal point after it in order to have five digits of the num- ber to the left of the decimal point. Hence x = 69,330. Ans. Ex. 14. Find the numbers whose logarithms are (a) 1.8055, (b) 1.4487, (c)0.2164, (d) 2.9487, (e)2.0529, (f) 5.2668, (g) 3.9774, (h)4.0010, (i) 8.4430 -10,* (j) 9.4975 - 10. Ans. (a) 63.9, (b) .281, (c) 1.646, (d) 888.6, (e) .011295, (f) 184,850, (g) 9493, (h) .00010023, (i) .02773, (j) .3144. * By (A), (B), p. 126, 8.4430-10= 2.4430. THEORY AND USE OF LOGARITHMS 135 70. The use of logarithms in computations. The following examples will illustrate how logarithms are used in actual calculations. Ex. 1. Calculate 243 x 13.49, using logarithms. Solution. Denoting the product by x, we may write x = 243 x 13.49. Taking the logarithms of both sides, we get log a; = log 243 + log 13.49. Th. I, p. 121 Looking up the logarithms of the numbers, log 243 = 2.3856 Rule II, p. 130 log 13.49 = 1.1300 Rule I, p. 129 Adding, logx = 3.5156 By Rule IV, p. 133, x = 3278. Ans. Ex. 2. Calculate Solution. Let x = 76,420 1375 x .06423 76,420 Then log x = log 1375 + log .06423 - log 76,420 Th. I, p. 121, and Th. II, p. 122 logl375= 3.1383 ' Rule I, p. 129 log. 06423 = 8.8077 - 10 Rule III, p. 132 Adding, 11.9460 - 10 log 76,420 = 4.8832 Rule III, p. 132 Subtracting, logx= 7.0628-10 or, logx= 3.0628. By Rule IV, p. 133, x = .0011555. Ans. Ex. 3. Calculate (5.664)». Solution. Let x = (5.664) 3 . Then log x = 3 log 5.664. Th. Ill, p. 122 log 5. 664 = 0. 7531 Rule III, p. 132 Multiplying by 3, § logx = 2.2593 By Rule IV, p. 133, x = 181.67. Ans. Ex. 4. Calculate -v^.7182. Solution. Let * = "8^7182 = (.7182)*. Then logx = J log. 7182. Th. IV, p. 122 log .7182 = 1.8662 Rule III, p. 132 = 29.8562-30. (b), Case IV, p. 128 Dividing by S, 3 129.8562-30 logx =9.9521 -10 = 1.9521. By Rule IV, p. 133, x = .8956. Ans. 136 PLANE TRIGONOMETRY Ex. 5. Calculate Solution. Let Then Dividing by 2, 'I V7194 x~87 "N 98,080,000 ' i= N 98,080,000 |~ (7194)*x 87 ")* " L 98,080,000 J ' Adding, log a; = ^ [£ log7194 + log 87 - log 98,080,000], log 7 194 = 3.8569 2 1 3. 8569 I log 7 194 = 1.9285 log 87 = 1.9395 3.8680 or, Subtracting, Dividing by 3, 13.8680 - 10 log 98,080,000 = 7.9916 5.8764 - 10 25.8764 - 30 3 |25.8764-30 logx = 8.6255- 10 = 2.6255. .-. x = .04222. Ans. (a), Case II, p. 128 (b), Case IV, p. 128 Ex. 6. Calculate Solution. Let 8 x 62.' '3 x .052 56 x 8.793 8 x = — x 62.73 x .052 Then 50 x 8.793 logx = [log 8 + log 62.73 + log. 052] - [log 56 + log 8.793]. log 8= 0.9031 log 62. 73= 1.7975 log .052 = 8.7160-10 log numerator = 11.4166 — 10 log denominator = 2.C924 logz= 8.7242-10 = 2.7242. .-. x = .05299. Ans.* log 56= 1.7482 log 8.793 = 0.9442 log denominator = 2.6924 * Instead of looking up the logarithms at once when we write down log 8, log 62.73, etc., it is better to write down an outline or skeleton of the computation before using the tables at all. Thus, for above example, log 8= 0. log 56=1. log 02.73= 1. log 8.793=0. log .062= 8. -10 log denominator = log numerator = log denominator = logx = .\ x = Tt saves time to look up all the logarithms at once, and, besides, the student is not so apt to forget to put down the characteristics. THEORY AND USE OF LOGARITHMS 137 71. Cologarithms. The logarithm of the reciprocal of a number is called its cologarithm (abbreviated colog). Hence if N is any positive number, colog N = log — = log 1 - log N Th. II, p. 122 = — log N = — log N, That is, the cologarithm of a number equals minus the logarithm of the number, the minus sign affecting the entire logarithm, both characteristic and mantissa. In order to avoid a negative mantissa in the cologarithm, it is customary to subtract the logarithm of the number from 10 — 10. Thus, taking 25 as the number, colog 25 = log ^ 3 = log 1 — log 25. But log 1 = 0, or, what amounts to the same thing, log 1 = 10.0000 - 10. Also, log 25 = 1. 3979 colog 25= 8.6021-10 Since dividing by a number is the same as multiplying by the reciprocal of the number, it is evident that when we are calculating by means of logarithms we may either subtract the logarithm of a divisor or add its cologarithm. When a computation is to be made in which several factors occur in the denominator of a fraction, it is more convenient to add the cologarithms of the factors than to sub- tract their logarithms. Hence Rule V. Instead of subtracting the logarithm of a divisor, we may add its cologarithm. The cologarithm of any number' is found by subtracting its logarithm from 10.0000 — 10. Ex. 1. Find colog 52. 63. Solution. 10.0000 - 10 log 52.63= 1.7212 colog 52. 63= 8.2788-10. Ans. Eule V Ex. 2. Find colog. 016548. Solution. 10.0000 - 10 log .016548 = g.2187 - 10 colog. 016548= 1.7813. Ans. Rule V Thus we see that the cologarithm may be obtained from the loga- rithm by subtracting the last significant figure of the mantissa from 10 and each of the others from 9. 138 PLANE TRIGONOMETRY In order to show how the use of cologarithms exhibits the written work in more compact form, let us calculate the expression in Ex. 6, namel y> 8 x 62.73 x -052 X 56 x 8.793 Solution. Using cologarithms, logx = log 8 + log 62.73 + log .052 + colog 56 + colog 8.793. log 8= 0.9031 log 62. 73= 1.7975 log. 052= 8.7160-10 colog 56= 8.2518-10 since log 56 = 1.7482 colog 8. 7 93 = 9.0558 - 10 since log 8. 793 = 0. 9442 logx = 28.7242 -30 = 2.7242. .-. x = .05299. Ans. Calculate the following expressions, using logarithms : 3. 9.238 x .9152. Ans. 8.454. n1 /.08726\* . „,„ 4.336.8 + 7984. .04218. Ul Umj " 4 «--«> la 5. (.07396)5. .000002213. 12. (5 38.2 x .000 5969)*. .8678. e _ 15.008x0843 ^ ^ u/TsTMyi i7//31.63\ 3 \ll29-j- .06376 x 4.248 7. V2. 1.414. 8. 1/ 5. 1.495. 14 - [w) 9. ^.02305. .2846. 15 . ^ x ^ x VM. 14. (» V. .6443. .7036. 10 . J^ 6 . .1606. 16. 1^1* -7.672. 962 52.37 17 18. (-2563) x .03442 An. .2415. 714.8 x (-.511) 121.6 x(- 9.025) _ Q725 (- 48.3) x 3662 x (- .0856) 72. Change of base in logarithms. We have seen how the logarithm of a number to the base 10 may be found in our tables. It is some- times necessary to find the logarithm of a number to a base different from 10. For the sake of generality let us assume that the loga- rithms of numbers to the base a have been computed. We wish to find the logarithm of a number, as N, to a new base b ; that is, we seek to express \og b N in terms of logarithms to the base a. Suppose log 6 iV = x, that is, b x = N. * From the definition of a logarithm, p. 119, it is evident that a negative number can have no logarithm. If negative numbers do occur in a computation, they should be treated as if they were positive, and the sign of the result determined by the rules for signs in Algebra, irrespective of the logarithmic work. Thus, in Example 16 above, we calculate the value' of 401.8 -f- 52.37 and write a minus sign before the result. THEORY AND USE OF LOGARITHMS 139 Taking the logarithms of both sides of this equation to the base a, log a fr T = log AT, or , x log„6 = log„JV. Th. Ill, p. 122 Solving, x = l p^. But log 6 N = x. By hypothesis. (90) ■•• log 6 * = 10g a AT log»& Theorem VI. The logarithm of a number to the new base b equals the logarithm, of the same number to the original base a, divided by the logarithm of b to the base a. This formula is also written in the form log„2\r = 7l/.log„iV, where M = is called the modulus of the new system with respect to the original one* This number M does not depend on the particular number N, but only on the two bases a and I. In actual computations a = 10, since the tables we use are com- puted to the base 10, Ex. Find logs 21. Solution. Here N = 21, b = 3, a = 10. Substituting in (90), log 3 21 = ^U« = 2.771. Ans. logi 3 .4771 EXAMPLES 1. Verify the following : (a) log 2 7 = 2.807. (e) log,, 8 = 0.0464. (i) logs 10 =2.096. (b) logs 4 = 1.262. (f) log 8 5 = 0.7740. (j) log 5 100 = 2.86. (o) log 4 9=1.585. (g) log, 14 = 1.366. (k) log„.l= -2.096. (d) log 6 7 = 1.209. (h) logj 102 = 2.873. (1) log 5 .01= -2.86. 2. Find the logarithm of T 7 T in the system of which 0.5 is the base. 3. Find the base of the system in which the logarithm of 8 is %. 4. Prove logs a ■ log„6 = 1. 1 5. Prove logjylO = logioiV * If, then, we have given the logarithms of numbers to a certain base a, and we wish to find the logarithms of the same numbers to a new base b, we multiply the given logarithms by the constant multiplier (modulus) .V== T - Thus, having given the common loga- rithms (base 10) of numbers, wc can reduce them to the logarithms of the same numbers to the base e(= 2.718) by multiplying them by M= = 2.3020. 140 PLANE TKIGONOMETEY 73. Exponential equations. These are equations in which the un- known quantities occur in the exponents. Such equations may often be solved by the use of logarithms, as illustrated in the following examples : Ex. 1. Given 81* = 10 ; find the value of x. Solution. Taking the logarithms of both members, log 81* = log 10, or, a; log 81 = log 10. Th. Ill, p. 122 , . log 10 1.0000 . _ ft . . Solving, x = — — = = 0.524. Ans. s ' log 81 1.9085 Ex. 2. Express the solution of a 1x + S x = c in terms of logarithms. Solution. Taking the logarithms of both members, l0g a 2a; + S _|_ log ft* = logc. Th. I, p. 121 (2 x + 3) log a + x log 6 = log a Th. Ill, p. 122 2 x log a + 3 log a + x log b = log c. x (2 log a + log 6) = log c — 3 log a. jE = log e -81oga j Ans 2 log a + log 6 Ex. 3. Solve the simultaneous equations (A) 2* • 3 v = 100. (B) x + y = 4. Solution. Taking the logarithms of both members of (A), and multiplying (B) through by log 2, we get x log 2 + y log 3 = 2 Th. I, III, p. 122 x log 2 + y log 2 = 4 log 2 Subtracting, y (log 3 — log 2) = 2 — 4 log 2 2-41og2 2-1.2040 Solving, y = log3-log2 .4771 -.3010 • 796 ° A KO y = = 4.52. .1761 Substituting back in (B), we get a; = — .52. EXAMPLES 1. Solve the following equations : (a) 5* = 12. Ans. 1.54. (g) (1.3)* =7.2. Ans. 7.53. (b) 7* = 25. 1.65. i l (c) (0.4)-* =7. 2.12. (h) (0.9)* 2 = (4.7) s. 0.45. (d) 10*- 1 = 4. 1.602. (i) 7*+« = 5. -2.1729. (e) 4*-! = 5*+!. - 13.43. (j) 22*+s-6*-i = 0. 9.5414 (f) 4* = 40. 2.66. THEORY AND USE OF LOGARITHMS^ 141 2. Solve the following simultaneous equations : (a) 4* ■ Zv = 8, Ans. x = .9005, (o) 2* • 2v = 2 22 , 4ns. x = 13, 2*-8<' = 9. ?/ = .7565. x - y = i. y = Q. (b) 3* • 4» = 15,552, x = 5, (d) 2* • 3» = 18, x = 1, 4* ■ 5* = 128,000. y = 3. 5* • 7? = 245. y = 2. 3. Indicate the solution of the following in terms of logarithms : (a) A = P(r + 1)*. Ans. x = ]2iAsL^I. log(r + l) (b)a*+»« = fc x = _l ± &. \ log a (c) a 1 • by = m, x : logc log d log m — log 6 log n log a log <J — log 6 log c c*-di< = ji. ^ logalogn -logc log m log a log d — log 6 log c (d) a 2;c - 3 -a 8 i'- 2 = a 8 , x=5, 3x + 2y=17. j/ = l. 74. Use of the tables of logarithms of the trigonometric functions. On p. 9 the values of the trigonometric functions of angles from 0° to 90° were given in tabulated form. When we are using loga- rithms in calculating expressions involving these trigonometric func- tions it saves much labor to have the logarithms of these functions already looked up for us and arranged in tabulated form.* Two com- plete sets of such logarithms of the trigonometric functions are given. Table II, pp. 8-16, should be used when the given or re- quired angle is expressed in degrees, minutes, and the decimal part of a minute ; and Table III, pp. 20-37, when the given or required angle is expressed in degrees, and the decimal part of a degree. - ) - In both tables the following directions hold true : Angles between 0° and 45° are in the extreme left-hand column on each page,! and the logarithm of the function of any angle will be found in the same horizontal row with it and in the vertical col- umn with the name of the function at the top ; that is, sines in the first column, tangents in the second, cotangents in the third, and cosines in the fourth, counting from left to right. * To distinguish between the two kinds of tables, that on p. 9 is called a Table of Natural Functions, while the logarithms of these functions arranged in tabulated form is called a Table of Logarithmic Functions. t The division of the degree into decimal parts, instead of using minutes and seconds, has much to recommend it theoretically, and is also regarded vith favor by many expert computers. In fact, a movement towards the adoption of such a system of subdivision is not only gaining headway in France and Germany, but is making itself felt in America. J The angles increase as we read downwards. 10 = 1.7941. p. 16 = 2.5363. p. 8 = 0.5905. p. 31 10 = 1.9994. p. 25 142 PLANE TRIGONOMETRY Angles between 45° and 90° are in the extreme right-hand column on each page,* and the logarithm of the function of any angle will be found in the same horizontal row with it and in the vertical col- umn with the name of the function at the bottom ; that is, cosines in the first column, cotangents in the second, tangents in the third, and sines in the fourth, counting from left to right. In order to avoid the printing of negative characteristics, the number 10 has been added to every logarithm in the first, second, and fourth columns (those having log sin, log tan, and log cos at the top). Hence in writing down any logarithm taken from these three columns — 10 should be written after it. Logarithms taken from the third column, having "log cot" at the top, should be used as printed. Thm - ' log sin 38° 30' = 9.7941 log cot 0° 10' = 2.5363 log tan 75.6° = 0.5905 log cos 2.94° = 9.9994 75. Use of Table II, pp. 8-i6, the given or required angle being expressed in degrees and minutes, f This table gives the logarithms of ' the sines, cosines, tangents, and cotangents of all angles from 0° to 5° and from 85° to 90° for each minute on pp. 8-12 ; and on pp. 13-16, from 5° to 85° at intervals of 10 minutes. The small columns headed "diff. V" immediately to the right of the columns headed " log sin " and " log cos " contain the differences, called tabular differences, in the logarithms of the sines and cosines corresponding to a difference of 1' in the angle. Similarly, the small column headed "com. diff. V" contains the tabular differences for both tangent and cotangent corresponding to a difference of 1' in the angle. It will be observed that any tabular difference is not in the same horizontal row with a logarithm, but midway between the two particular logarithms whose difference it is. Of course that tabular difference should always be taken which corresponds to the interval in which the angle in question lies. Thus, in finding log cos 78° 16', the tabular difference corresponding to the interval between 78° 10' and 78° 20' is 6.1. * The angles increase as we read upwards. t In case the given angle involves seconds, first reduce the seconds to the decimal part of a minute by dividing by 60. Thus, 88° 18' 42" = 88" 18.7', since 42" <= $$ = .7' ; 2° 0' 16" = 2" 0.27', since 16" = \§ = .266' If the angle is given in degrees and the decimal parts of a degree, and it is desired tc use Table II, the angle may be quickly found in degrees and minutes by making use of the Conversion Table on p. 17. THEORY AND USE OF LOGARITHMS 143 76. To find the logarithm of a function of an angle when the angle is expressed in degrees and minutes, use RuiiE VI. When the given angle is found exactly in Table II, the logarithm of the given function of the angle is immediately found in the same horizontal row and in the vertical column having the given function at the top when the angle is less than 45", or at the bottom when the angle is greater than 45°. In case the given angle is not found exactly in the table we should take the following steps : (a) Write down the logarithm of the same function of the next less angle found in the table, and also the corresponding tabular difference for 1'. (b) To find the correction necessary, multiply- this tabular differ- ence by the excess in minutes of the given angle over the angle whose logarithm was written down. (c) If sine or tangent, add } , , . , . ^ Z-. . . f- this correction.* If cosine or cotangent, subtract J This rule, as well as the next three, assumes that the differences of the logarithms of functions are proportional to the differences of their corresponding angles. Unless the angle is very near 0° or 90°, this is in general sufficiently exact for most practical purposes. Ex. 1. Find log tan 32° 30'. Solution. On p. 15, Table II, we find the angle 32° 30' exactly ; hence, by Eule VI, we get immediately from the table log tan 32° 30' = 9.8042 - 10. Ans. Ex. 2. Find log cot 88° 17'. Solution. On p. 9, Table II, we find the angle 88° 17' exactly ; hence, by Kule VI, we get at once log cot 88° 17' =8.4767 -10. Ans. Ex. 3. Find log sin 23° 26'. Solution. The exact angle 23° 26' is not found in Table II ; but then, by Rule VI, from p. 14, log sin 23° 20' = 9. 5978 - 10 Tab . mtt . = 2 .9 corr. for 6' = 17 Excess = 6 log sin 23° 26' = 9. 5995 -10. Ans. Corr - =rM * The sine and tangent increase as the angle increases, hence we add the correction ; the cosine and cotangent, however, decrease as the angle increases, hence we Bubtract the correction. Of course this is true only for acute angles. 14-1 PLANE TRIGONOMETRY Ex. 4. Find log cos 54° 42' 18". Solution. Since 18" is less than half a minute, we drop it, and from p. 16, Table II, by Rule VI, log cos 54° 40' = 9. 7622 - 10 Tab - diff - = L8 j. „, , Excess = 2 corr. for 2' = 4 Corr = £ g logcos54°42'= 9.7618 -10. Ans. l.e.=4 Ex. 5. Find log cot 1° 34.42'. Solution. From p. 9, Table II, by Rule VI, log cot 1° 34' =1.6630 Tab.diff.= 46 corr. for .4' = 18 Excess = .4 log cot 1° 34.4' =1.5612. Ans. Corr - = 18 - 4 When the angles are given in the table at intervals of 10', it is only necessary to take our angle to the nearest minute, while if the angles are given for every minute, we take our angle to the nearest tenth of a minute. Thus, in Ex. 4, we find cos 54° 42', dropping the seconds ; and in Ex. 5 we find log cot 1° 34.4', dropping the final 2. Ex. 6. Verify the following : (a) log tan 35° 50' = 9.8586 - 10. (g) log cos 27° 28' = 9.9480 - 10. (b) logsin61° 58' =9.9458 -10. (h) log cot 61° 49' = 9.8957 - 10. (c) log tan 82° 3' 20"= .8550. (i) log sin 85° 57' = 9.9989 - 10. (d) log cos 44° 32' 50"= 9.8528 -10. (j) log cot 45° 0' 13" = 0.0000. (e) log tan 1° 53.2' = 8.5178 - 10. (k) log sin 120° 24.3' = 9.9358 - 10. (i) log tan 87° 15.6' =1.3201. (1) log tan 243° 42' 16" = 0.3060. 77. To find the acute angle in degrees and minutes which corresponds to a given logarithmic function, use Rule VII. When the given logarithmic function is found exactly in Table II, then the corresponding angle is immediately found in the same horizontal row, to the left if the given function is written at the top of the column, and to the right if at the bottom. In case the given logarithmic function is not found exactly in the table we should take the following steps : (a) Write down the angle corresponding to the next less logarithm of the same function found in the table, and also the corresponding tabular difference for 1'. (b) To find the necessary correction in minutes divide this tabular difference into the excess of the given logarithmic function over the one written down. (c) If sine or tangent, add 1 , . . * ' . > this correction.* If cosine or cotangent, subtract J * See footnote, p. 143. THEOEY AND USE OF LOGARITHMS 145 In searching the table for the given logarithm, attention must be paid to the fact that the functions are found in different columns according as the angle is less or greater than 45°. If, for example, the logarithmic sine is found in the column with "log sin" at the top, the degrees and minutes must be taken from the left-hand. column, but if it is found in the column with " log sin " at the bottom, the degrees and minutes must be taken from the right-hand column. Similarly, for the other functions. Thus, if the logarithmic cosine is given, we look for it in two columns on each page, the one having " log cos " at the top and also the one having " log cos " at the bottom. Ex. 7. Find the angle whose log tan = 9.6946 — 10. Solution. This problem may also be stated as follows : haying given log tan x = 9.6946 — 10 ; to find the angle x. Looking up and down the columns having " log tan " at top or bottom, we find 9.6946 exactly on p. 15, Table n, in the column with " log tan " at top. The corresponding angle is then found in the same horizontal row to the left and is x = 26° 20'. Ex. 8. Find the angle whose log sin = 9.6652 — 10. Solution. That is, having given log sinx = 9.6652 — 10 ; to find the angle x. Looking up and .down the columns having " log sin " at top or bottom, we do not find 9.6652 exactly ; but (Rule VII) the next less logarithm in such a column is found on p. 15, Table II, to be 9.6644, which corresponds to the angle 27° 30 / , and the corresponding tabular difference for 1' is 2.4. Hence log sins = 9.6652 — 10 Tab. dlff.l'l Excess |Corr. log sin 27° 30" = 9.6644 - 10 — 72 excess = ° 8 Since the function involved is the sine, we add this correction, giving x = 27° 30' +3' =27° 33'. Ant. Ex. 9. Find the angle whose log cos = 9.3705 — 10. Solution. That is, having given log cos x = 9.3705 — 10 ; to find the angle x. Looking up and down the columns having " log cos " at top or bottom, we do not find 9.3705 exactly ; but (Rule VII) the next less logarithm in such a col- umn is found on p. 13, Table II, to be 9.3682, which corresponds to the angle 76° 3<y, and the corresponding tabular difference for V is 5.2. Hence logcosx = 9.3705 - 10 Tab.difl.riExcesslCorr. log cos 76° 30' = 9.3682 - 10 — ' ^? L-^— excess = 23 ■:i Since the function involved is the cosine, we subtract this correction, giving x = 76° 30' -4' = 76° 26'. Ans. 146 PLANE TRIGONOMETRY Ex . 10. Given log tan x = 8. 7670 - 10 ; find x. Solution. By Rule VII the next less logarithmic tangent is found on p. 11, Table II. logtanx = 8.7570 - 10 Tab. diff.l' [Excess! Corr. 22 5.0 .2 log tan 3° 16' = 8.7665-10 excess = 5 Hence x = 3° 16' + .2' = 3° 16.2'. Ans. 44 6 13.0 Ex. 11. Given cotx = (1.01) 6 ; find x. Solution. Taking the logarithms of both sides, log cot x= 5 log 1.01. Th. Ill, p. 122 But log 1.01 = 0.0043 and, multiplying by 5, 5 log cotx = 0.0215 ; to find x. By Rule VII the next less logarithmic cotangent is found on p. 16, Table II. log cotx = 0.0215 Tab.diff.l' [Excess! Corr. log cot 43° 40' = 0.0202 2.6 \ 13.0 | 5 excess = 13 Hence x = 43° 40' - 5' = 43° 36'. Ans. Ex. 12. Verify the following : (a) H log sin x = 9.5443 - 10, then x = 20° 30'. (b) If log cosx = 9.7531 - 10, then x = 55° 30'. (c) If log tan x = 9.9570 - 10, then x = 42° 10'. (d) If log cotx = 1.0034, then x = 5° 40'. (e) If log sin x = 8.0436 - 10, then x = 0° 38'. (f) If log cosx = 8.7918 - 10, then x = 86° 27' (g) If log tan x = 9.5261 - 10, then x = 18° 34'. (h) If log cotx = 0.6380, then x = 12° 58'. (i) If log sin x = 9.9995 - 10, then x = 87° 16'.* (j) If log cosx = 8.2881 - 10, then x = 88° 53.3'. (k) If log tan x = 2. 1642, then x = 89° 36.4'. _ (1) If log tanx = 7.9732 - 10, then x = 0°32.3'. (m) If log sin x = 9.8500 - 10, then x = 45° 4'. (n) If log cos x = 9.9000 - 10, then x = 37° 25'. (o) If log tan x = 0.0036, then x = 45° 14'. (p) If log cotx = 1.0000, then x = 5° 43'. (q) If log cotx = 3.9732, then x = 89° 27.7'. * "When there are several angles corresponding to the given logarithmic function, we choose the middle one. THEORY AND USE OF LOGARITHMS 141 EXAMPLES Use logarithms when making the calculations in the following examples : 1. Given 184sinS x = (12.03)2 cos 57° 20' ; find x. Solution. First we solve for sin j;. giving „„,._ » /(ia.08}» eoe57° ay V 184 Taking the logarithms of both sides, log sin x = ^ [2 log 12.03 + log cos 57° 20' + colog 184]. 2 log 12.03= 2.1606 since log 12.03 = 1.0803 log cos 57° 20' = 9. 7322 - 10 colog 184 = 7.7352 - 10 since los* 184 = 2.2648 19.6280 -20 3 |29.0280 -30 log sin j = 9.8700 -10 . I = 4S 3 44'. -las. 2. Given cosi = (.9854)*; find x. 3 Calcnl k 4.236 cos 52° ate 1{K Ans. 5^45'. " 13.087 sin 48^ 5' " 26 ° 9- 4. Given 1.5 cot 82 D = x- sin 12° 15'; find x. .9968. Bint. First solve for x, giving ' i j cot sa° _ \ sin li" 15' ' 5. Given 50 tan i = -^.2584 ; find x. 0° 49'. „ _ , , . sin 24° 13' cot 58° 2" 6. Calculate 84-ir cos 33° 17' tan 19° 58' ' " " 7. Calculate Vcos 10° 5' ta n 73° 11'. 1.805. „„,,<. (sin33°18'V<Vcot71°20' 8. Calculate 0044°° 10.658 tan 63° 54' ■""-«— 9. Given 3 cot x = ¥7; ; find x. 72° 45'. 10. Given sinz = (.9361) 10 ; find jr. 31° 6'. 11. Given 2.3 tan x = (1.002) la5 ; find x. 29P 24'. 78. Use of Table HI, pp. 20-37, the given or required angle being expressed in degrees and the decimal part of a degree.* This table gives, on pp. 20-29, the logarithms of the sines, cosines, tangents, and cotangents of all angles from 0° to 5°. and from 85° to 90° for every hundredth part of a degree ; and on pp. 30-37 from 5° to 85° for every tenth of a degree. The tabular differences between the logarithms given in the table are given in the same manner as were the tabular differences in Table II, and the general arrangement is the same. ■ In case the angle is given in degrees, minutes, and seconds, and it is desired to use Table TTX, we may quickly reduce the angle to degrees and the decimal part of a degree by using the Conversion Table on p. 17. 148 PLANE TRIGONOMETRY 79. To find the logarithm of the function of an angle when the angle is expressed in degrees and the decimal part of a degree, use Rule VIII. When the given angle is found exactly in Table III, the logarithm of the given function of the angle is immediately found in the same horizontal row and in the vertical column having the given function at the top when the angle is less than J/.5°, or at the bottom when the angle is greater than 45°. In case the given angle is not found exactly in the table we should take the following steps : (a) Write down the logarithm of the same function of the next less angle * found in the table and note the tabular difference which follows. (b) In the Prop. Parts column locate the block corresponding to this tabular difference. Under this difference and opposite the extra digit of the given angle will be found the proportional part of the tabular difference (that is, the correction). (c) If sine or tangent, add 1 , . . v ' z. > this correction.^ If cosine or cotangent, subtract ) Ex. 1. Find log sin 27. 4°. Solution. On p. 34, Table III, we find the angle 27.4° exactly; hence, by Rule VIII, we get at once log sin 27.4° = 9.6629 - 10. Ana. Ex. 2. Find log cot 3. 17°. Solution. On p. 26, Table III, we find the angle 3.17° exactly; hence, by Rule VIII, we get immediately from the table log cot 3. 17° = 1.2566. Ans. Ex.3. Find log tan 61.87°. Solution. The exact angle 61.87° is not' found in our tables. But then, by Rule VIII, the next less angle is 61.8°, the extra digit of the given angle being 7, and we have, from p. 34, Table III, log tan 61.8° = 10.2707 - 10. The tabular difference between log tan 61.8° and log tan 61.9° is 18. In the Prop. Parts column under 18 and opposite the extra digit 7 we find the pro- portional part 12. 6 ( = 13). Then log tan 61. 80° =0.2707 13 Prop. Part. log tan 61.87° = 0.2720. Ana. * This "next less angle" will not contain the last (extra) digit of the given angle. ■t See footnote, p. 143. THEORY AND USE OF LOGARITHMS 149 Ex. 4. Find log cot2.158°. Solution. The exact angle 2.158° is not found in our tables. But then, by- Rule VIII, the next less angle is 2. 15°, the extra digit of the given angle being 8, and we have, from p. 24, Table III, log cot 2. 15° =1.4255. The tabular difference between log cot 2. 15° and log cot 2. 16° is 20. In the Prop. Parts column under 20 and opposite the extra digit 8 we find the propor- tional part 16. Then log cot 2. 150° =1.4255 16 Prop. Part, log cot2.158° = 1.4239. Ann. Ex. 5. Verify the following : (a) log tan 37.6° = 9.8865 - 10. (g) log tan 88.564° = 1.6009. (b) log sin 63.87° = 9.9532 - 10. (h) log cos20.03° = 9.9729 - 10. (c) log cot 1.111° = 1.7123. (i) log sin 89.97° = 0.0000. (d) log sin 0.335° = 7.7669 - 10. (j) log cot34.84° = 0.1574. (e) log cos45.68° = 9.8443 - 10. (k) log sin 155.42° = 9.6191 - 10. (f) log tan 3.867° = 8.8299 - 10. (1) log tan 196.85° = 9.4813 - 10. 80. To find the acute angle in degrees and decimal parts of a degree which corresponds to a given logarithmic function, use Rule IX. When the given logarithmic function is found exactly in Table III, then the corresponding angle is immediately found in the same horizontal row ; to the left, if the given function is written at top of the column, and to the right if written at the bottom. In case the given logarithmic function is not found exactly in the table we should take the following steps : (a) Locate the given logarithm between two of the logarithms of the same function given in the tables. (b) The lesser angle of the two angles corresponding to these logarithms will be the required angle complete except for the last digit. Write this angle down with the corresponding logarithmic function. (c) Find the difference between the logarithm just written down and the given logarithm, also noting the corresponding tabular difference in the table. (d) In the Prop. Parts column, under this tabular difference, pick out the proportional part nearest the difference found in (c), and to the left of it will be found the last (extra) digit of the required angle, which we now annex. 150 PLANE TRIGONOMETRY Ex. 6. Having given log tan x = 9. 5364 — 10 ; to find the angle x. Solution. Looking up and down the columns having "log tan" at top oi bottom, we do not find 9.5364 exactly. But then, by liule IX, we locate n between 9.5345 and 9.5370, on p. 32, Table III. Except for the last digit the required angle will be the lesser of the two corresponding angles, that is, 18.9°. Then log tan 18. 9° =9. 6345 -10 log tan x = 9.5364-10 19 = difference. The corresponding tabular difference being 25, we find in the Prop. Parts col- umn that 20 is the proportional part under 25 which is nearest 19. To the left of 20 is the last (extra) digit 8 of the required angle. Hence x = 18.98°. Ans. Ex. 7. Having given log cosx = 8.6820 — 10 ; find x. Solution. On p. 25, Table III, we locate 8.6820 between 8.6810 and 8.6826. Except for the last digit, the required angle must be the lesser of the.two cor- responding angles, that is, 87.24°. Then log cos 87.24° = 8.6826 - 10 log cosx = 8.6820 - 10 6 = difference. The corresponding tabular difference being 16, we find in the Prop. Parts col- umn that 6.4 is the proportional part under 16 which is nearest 6. To the left of 6.4 is the last (extra) digit 4 of the required angle. Hence x = 87.244°. Ans. Ex. 8. Verify the following : (a) If log sin x = 9.6371 - 10, then x = 25.7°- (b) If log cosx = 9.9873 - 10, then x = 13.8°. (c) If log tanx = 8.9186 - 10, then x = 4.74°. (d) If log cot x = 1.1697, then x = 3.96°. (e) If log sinx = 9.5062 - 10, then x = 18.67° (f) If log cosx = 9.9629 - 10, then x = 23.35°. (g) If log tanx = 9.8380 - 10, then x = 34.55°. (h) If logcotx = 9.3361 - 10, then x = 77.77°. (i) If log sinx = 8.6852 - 10, then x = 2.776°. (j) If log cos x = 9.9995 - 10, then x = 2.74°. (k) If log tanx = 7.2642 - 10, then x = 0.105°. (1) If log cotx = 1.7900, then x = 0.929°. (m) If log sinx = 9.5350 - 10, then x = 20.05°. (n) If log cosx = 9.8000 - 10, then x = 50.88°. (o) If log tanx = 0.0035, then x = 45.23°. (p) If log cotx = 2.0000, then x = 0.573°. (q) If log sinx = 0.0000, then x = 90°. (r) If log tan x = 0.0000, then x = 45°. THEORY AND USE OF LOGARITHMS 151 EXAMPLES Use logarithms when making the calculations in the following examples: 1. Given tan x = (1.018)!2; find x. Solution. Taking the logarithms of both sides, log tan x = 12 log 1.018. Th. Ill, p. 122 But and, multiplying by 12, log 1.018 = 0.0077 12 log tanx = 0.0924 On p. 36 we locate 0.0924 between 0.0916 and 0.0932. Then log tan 51.0° =0.0916 log tan x = 0.0924 8 = difference. The tabular difference is 16. In the Prop. Parts column under 16 we find 8.0 exactly. To the left of 8.0 we find the last digit 5 of the required angle. Hence x = 51.05°. Ans. 2. Given 56.4 tan 5 x = (18.65) 5 cos69.8°; finds. Solution. First we solve for tan x, giving 5/(18.65)5 cos 69.8° tan x = \ : \ 56.4 Taking the logarithms of both sides, log tan a; = \ [5 log 18.65 + log cos69.8°+ colog 56.4]. 5 log 18.65= 6.3535 log cos 69. 8°= 9.5382 -10 colog 56.4 = 8.2487 - 10 24.1404-20 5 |54.1404-50 log tan x = 10.8281 - 10. .-. x = 81.55°. Ans. 3. Given cosx =V.9681; find x. , , 26.52 tan 33.86° 4 - Calculate i0086eot88g63O - 5. Given V§ sin 48.06° = x 8 cos 2. 143° ; find X. Hint. First solve for x, giving cos 2.143" _ J 1 /VI sin 48.06° 6. Given 5 cot x = ^.4083 ; find x. V83 cos 52.82° 7. Given sin x ■■ (13. 382) 2 - ; find X. 8. Calculate V361 tan 87. 5° sin 9. 53° since log 18.65 = 1.2707 since log 56.4 = 1.7513 Ans. 10.25°. 9.745. 1.0885. 81.56°. 1.762°. 37. 152 PLANE TRIGONOMETRY 81. Use of logarithms in the solution of right triangles. Since the solutions of right triangles involve the calculation of products and quotients, time and labor may be saved by using logarithms in the computations. From p. 7 we have the following : General directions for solving right triangles. First step. Draw a figure as accurately as possible representing the triangle in question. Second step. When one acute angle is known, subtract it from, 90° to get the other acute angle. Third step. To find an unknown part, select from (1) to (6), p. 2, a formula involving the unknown part and two known parts, and then solve for the unknown part* Fourth step. Check the values found by seeing whether they satisfy relations different from those already employed in the third step. A convenient numerical check is the relation a 2 = c 2 - b 2 = (c + b) (p - S).f Large errors may be detected by measurement. For reference purposes we give the following formulas from p. 8 and p. 11. , t Area of a right triangle = — • (7) Side opposite an acute angle = hypotenuse x sine of the angle. (8) Side adjacent an acute angle = hypotenuse x cosine of the angle. (9) Side opposite an acute angle = adjacent side x tangent of the angle. It is best to compute the required parts of any triangle as far as possible from the given parts, so that an error made in determining one part will not affect the computation of the other parts. * This also includes formulas (7), (8)» (9). on p. 11. t When we want the hypotenuse, the other two sides being given, this formula is not ffell adapted to logarithmic computation, since c= Vaa + b% and we have a summation under the radical that cannot he performed hy the use of our logarithmic tables. If, however, we have the hypotenuse c and one side (as b) given to find the other side a, then a= Vc 2 -& 2 = V(c-6)(c + 6), and we have a product under the radical. The factors c-b and c + b of this product are easily calculated hy inspection, and then we can use logarithms advantageously. Thus log a = £ [log (c-b) + log (c + ft)]. t In case a or 6 is not given, or both a and 6 are not given, we first find what we need from the known parts, as when solving the triangle, so that we can use the above formula for finding the area. THEOEY AND USE OF LOGAEITHMS 153 In trigonometric computations it sometimes happens that the unknown quantity may be determined in more than one way. When choosing the method to be employed it is important to keep in mind the following suggestions : (a) An angle is best determined from a trigonometric function which changes rapidly, that is, one having large tabular differences, as the tangent or cotangent. (b) When a number is to be found (as the side of a triangle) from a relation involving a given angle, it is best to employ a trigonometric function of the angle which changes slowly, as the sine or cosine. As was pointed out on pp. 13, 14, the solution of isosceles triangles and regular polygons depends on the solution of right triangles. The following examples will illustrate the best plan to follow in solving right triangles by the aid of logarithms. Ex. 1. Solve the right triangle if A = 48° 17', c = 324. Also find the area. Solution. First step. Draw a figure of the triangle indicating the known and unknown parts. Second step. B=90°-A = 41° 43 > . Third step. To find a use a = c sin .A. Taking the logarithms of both sides, log a = log c + log sin A. Hence, from Tables I and II,* logc= 2.5105 log sin A - 9.8730 - 10 loga = 12.3835 -10 = 2.3835. .-. a = 241.8. To find 6 use b = c cos^i . Taking the logarithms of both sides, log 6 = log c + log cos A. Hence, from Tables I and II, logc= 2.5105 log cos A = 9.8231 - 10 log 6 = 12.3336 -10 = 2.3336 .-. 6 = 215.6. « If we wish to use Table III instead of Table II, we reduce 17' to the decimal of a degree. Thus, A= 48° 17' = 48.28°. 154 PLATTE TRIGONOMETRY Fourth step. To,check these results numerically, let us see if a, b, c satisfy the equation a 2 = c 2 - 6 2 = (c + 6) (c - 6), or, using logarithms, 2 log a =,log(c + 6) + log(c — 6), that is, log a = £ [log (c + 6) + log (c — 6)]. Here c + b = 639.6 and c - 6 = 108.4. log(c + 6) = 2.7321 log(c- 6) = 2.0350 2 log a = 4.7671 loga= 2.3835. Since this value of log a is the same as that obtained above, the answers are probably correct. To find the area use formula . ab Area = — 2 log area = log a + log b - log 2. loga = 2.3835 log 6 = 2.3336 4.7171 log 2 = 0.3010 log area = 4.4161 .-. area = 26,070. Ex. 2. Solve the right triangle, having given 6 = 15.12, c = 30.81. Solution. Here we first find an acute angle ; to find A use ooaA = -. (2), p. 2 log cos .4 = log 6 — log c. . log 6 = 11. 1796 -10 loge= 1.4887 logcos^= 9.6909-10 . . A = 60° 36'. from Table II, p. 15 B = 90° - A = 29° 24'. Hence 15.12 To find a we may use a = bt&nA. log a = log b + log tan .4. logo =1.1796. log tan 4. = 0.2491 log a = 1.4287 by (9), p. 11 a = 26.84. THEORY AND USE OF LOGARITHMS 155 To check the work numerically, take o*=(e + 6)(c-6), or > log a = £ [log (c + 6) + log (c - &)] . Here c + b = 45.93 and c-b = 16.69. log(c + 6) = 1.6621 log(c - 6) = 1.1956 2 log a = 2.8578 loga = 1.4288. This we see agrees substantially with the above result. Ex. 3. Solve the right triangle, having given B = 2.325°, a = 1875.3. Solution. A = 90° - B = 87.676° sin^ = -. by(l), p. 2 Solving for the unknown side c, _ a smA logo = loga — log sin A, Hence, from Tables I and III,* B log a = 13.2731- - 10 ii log sin A= 9.9996- - 10 C3 logc= 3.2735 .-. c = 1877. A ti c tan.4 = -• 6 by (3), p. 2 Solving for the unknown side b, b= a . tan A log b = log a — log tan A loga= 13.2731 -10 log tan A = 11.3915-10 log 6= 1.8816 .-. 6=76.13. To check the work we may use formulas a?=(c + b) (c - 6), or, b — csinB, by (7), p. 11 since neither one was used in the above calculations. * If -we wish to use Table II instead of Table III, we reduce 2.325° to degrees and min- utes. Thus, B= 2.325° = 2° 19.5'. 156 PLANE TEIGONOMETEY EXAMPLES Solve the following right triangles (C= 90°), using logarithmic Tables I and II. « No. Given Parts Req jired Parts 1 A = 43° 30' c = 11.2 B = 46° 30' a= 7.709 6 = 8.124 2 B = 68° 50' a = 729.3 A = 21° 10' 6= 1883.5 c = 2019.5 3 B = 62° 56' 6 = 47.7 A = 27° 4' a = 24.37 c = 53.56 4 a = .624 c = .91 A = 43° 18' B = 46° 42' 6 = .6623 5 A = 72° 7' a = 83.4 B = 17° 53' 6 = 26.91 c = 87.64 6 6 = 2.887 e = 5.11 B = 34° 24' A = 55° 36' a = 4.216 7 A = 52° 41' b = 4247 B = 37° 19' a = 5571 c = 7007 8 a = 101 6 = 116 A = 41° 2' 5 = 48° 58' c= 153.8 9 A = 43° 22' a = 158.3 £ = 46° 38' 6=167.6 c = 230.5 10 a = 204.2 c = 275.3 A = 47° 53' B = 42° 7' 6= 184.7 11 B = 10° 51' c = .7264 A = 79° 9' a = .7133 6 =.1367 12 a = 638.5 6 = 501.2 4 = 51° 53' B = 38° r c = 811.7 13 6 = .02497 c = .04792 A = 58° 36' B = 31° 24' a = .0409 14 B = 2° 19' 30" a = 1875.3 -4 = 87° 40' 30" 6=76.13 c = 1877 15 £ = 21° 33' 51" a = .8211 A = 68° 26' 9" 6 = .3245 c = .8829 16 ^1 = 74° 0' 18" c = 275.62 B = 15° .59' 4^' a = 264.9 6 = 75.95 17 B = 34° 14' 37" 6 = 120.22 A = 55° 45' 23" a =176.57 c = 213.6 18 a= 10.107 6 = 17.303 A = 30° 17.6' B = 59° 42.4' c = 20.04 19 a = 24.67 6 = 33.02 4 = 36° 46' B = 53° 14' c = 41.22 20 ^1 = 78° 17' = 203.8 B = 11° 43' 6 = 42.27 c = 208.15 21. Find areas of the first five of the above triangles. Ans. (1) 31.32; (2) 686,900; (3) 681.3; (4) .2067; (5) 1122.5. Solve the following isosceles triangles where A, B, C are the angles and a, b, c the sides opposite respectively, o and 6 being the equal sides. 22. Given A = 68° 57', 6 = 35.09. Ans. C= 42° 6', c = 25.21. 23. Given B = 27° 8', c = 3.088. Ans. C =125° 44', a = 1.735. 24. Given C = 80° 47', 5 = 2103. Ans. A = 49° 36.5', c = 2725.4. 25. Given a = 79.24, c = 106.62. Ans. A = 47° 43', C = 84° 34'. 26. Given C = 151° 28', c = 95.47. Ans. A = 14° 16', a = 49.25. 27. One side of a regular octagon is 24 ft. ; find its area and the radii of the inscribed and circumscribed circles. Ans. Area = 2782, r = 28.97, R = 31.36. * For the sake of clearness and simplicity, one set of triangle examples is given which are adapted to practice in using Table II, the given and required angles being expressed in degrees and minutes ; and another set is given on p. 157 for practice in the use of Table III, the given and required angles being expressed in degrees and the decimal part of a degree. There is no reason why the student should not work out the examples in the first set, using Table III, and those in the second set, using Table II, if he so desires, except that it may involve a trifle more labor. This extra work of reducing minutes to the decimal part of a degree, or the reverse, may be reduced to a minimum by making use of the Conversion Tables on p. 17. It is possible, however, that an answer thus obtained may differ from the one given here by one unit in the last decimal place. This practice of giving one set of triangle examples for each of the Tables II and III will be followed throughout this book when solving triangles. THEORY AND USE OF LOGARITHMS 157 Solve the following right triangles (C= 90°), using logarithmic Tables I and III. No. Given Parts Required Parts 28 a = 5 6 = 2 4 = 68.2° B = 21.8° c = 5.385 29 £ = 32.17° c = .02728 A = 57.83° a = .02309 6 = .01452 30 A = 58.65° c = 35.73 B = 31.35° a = 30.51 6=18.59 31 A = 22.23° 6 = 13.242 £ = 67.77° a = 6.413 c = 14.31 32 6 = .02497 c = .04792 A = 58.6° £ = 31.4° a = .0409 33 a = 273 6 = 418 4=33.15° £ = 56.85° c = 499.3 34 £ = 23.15° 6=75.48 A = 66.85° a =176.5 c = 191.9 35 A= 31.75° a = 48.04 B = 58.25° 6 = 77.64 c = 91.28 36 6 = 512 c = 900 A= 55.32° £ = 34.68° a = 740.2 37 a= 52 c = 60 A = 60.06° £ = 29.94° 6 = 29.94 38 A = 2.49° a = .83 £ = 87.51° 6 = 19.085 c = 19.107 39 4 = 88.426° 6 = 9 £= 1.574° a = 327.5 c = 327.6 40 £ = 4.963° 6 =.07 4 = 85.037° a = .8062 c = .8092 41 £ = 85.475° c = 80 4 = 4.525° a = 6.313 6 = 79.74 42 a = 100.87 6 = 2 4 = 88.864° £= 1.136° c = 100.9 43. Find the areas of the first five of the above triangles. 4ns. (28) 5; (29) .0001677; (30) 283.6; (31) 35.84; (32) .00051. 44. The perimeter of a regular polygon of 11 sides is 23.47 ft. Find the radius of the circumscribed circle. Ans. 3.79 ft. 45. Two stations are 3 mi. apart on a plain. The angle of depression of one from a balloon directly over the other is observed to be 8° 15'. How high is the balloon? 4ns. .435 mi. 46. A rock on the bank of a river is 130 ft. above the water level. From a point just opposite the rock on the other bank of the river the angle of elevation of the rock is 14° 30' 21". Find the width of the river. 4ns. 502.5 ft. 47. A rope 38 ft. long tied to the top of a tree 29 ft. high just reaches the level ground. Find the angle the rope makes with the tree. 4ns. 40° 15'. 48. A man 5 ft. 10 in. high stands at a distance of 4 ft. 7 in. from a lamp-post, and casts a shadow 18 ft. long. Find the height of the lamp-post. 4ns. 7.32 ft. 49. The shadow of a vertical cliff 113 ft. high just reaches a boat on the sea 93 ft. from its base. Find the altitude of the sun. 47is. 50° 33'. 50. The top of a tree broken by the wind strikes the ground 15 ft. from the foot of the tree and makes an angle of 42° 28' with the ground. Find the origi- nal height of the tree. 4ns. 34.07 ft. 51. A building is 121 ft. high. From a point directly across the street its angle of elevation is 65° 3'. Find the width of the street. 4ns. 56.3 ft. 52. Given that the sun's distance from the earth is 92,000,000 mi., and the angle it subtends from the earth is 32'. Find diameter of the sun. 4ns. About 856,400 mi. 53. Given that the radius of the earth is 3963 mi. , and that it subtends an angle of 57' at the moon. Find the distance of the moon from the earth. 4ns. About 239,017 mi. 158 PLANE TKIGONOMETEY 54. The radius of a circle is 12,732, and the length of a chord is 18,321. Find the angle the chord subtends at the center. Ans. 02° 2'. 55. If the radius of a circle is 10 in., what is the length of a chord which subtends an angle of 77° 17' 40" at the center ? Ana. 12.488 in. 56. The angle between the legs of a pair of dividers is 43°, and the legs are 7 in. long. Find the distance between the points. Ans. 5. 13 in. 82. Use of logarithms in the solution of oblique triangles. As has already been pointed out, formulas involving principally products, quotients, powers, and roots are well adapted to logarithmic compu- tation; while in the case of formulas involving in the main sums and differences, the labor-saving advantages of logarithmic compu- tation are not so marked. Thus, in solving oblique triangles, the law of sines a b o sin. A sin 12 sinC and the law of tangents tan I (A - B) = |^| tan i(A + B), are well adapted to the use of logarithms, while this is not the case with the law of cosines, namely, a 2 = 6 2 + c 2 — 2 be cos A. In solving oblique triangles by logarithmic computation, it is con- venient to classify the problems as follows : Case I. When two angles and a side are given. Case II. When two sides and the angle opposite one of them are given (ambiguous case). Case III. When two sides and included angle are given. Case IV. When all three sides are given. Case I. When two angles and a side are given. First step. To find the third angle, subtract the sum of the two given angles from 180°. Second step. To find an unknown side, choose a pair of ratios from the law of sines a b c smA sin .8 sinC which involve only one unknown part, and solve for that part. Check : See if the sides found satisfy the law of tangents. THEORY AND USE OE LOGARITHMS 159 Ex. 1. Having given 6 = 20, A = 104°, B = 19° ; solve the triangle. Solution. Drawing a figure of the triangle on which we indicate the known and unknown parts, we see that the problem comes under Case I. First step. C = 180° - (A + B) = 180° - 123° = 57°. Second step. Solving sin A sin 2J 6 sin A for a, we get a : or, sinB log a = log 6 + log sin A — log sin B. log 6= 1.3010 log sin A = 9. 9869 - 10 * 11.2879 - 10 log sin B = 9.5126 - 10 loga= 1.7753 a = 59.61. Solving sin B sin C for c, we get c = bsinC or, sin J? log c = log 6 + log sin C — log sin B. log 6= 1.3010 log sin C = 9.9236 10 Check : Here, 11.2246-10 log sin B = 9.5126-10 logc= 1.7120 c = 51.52. a + c = 111.13, a-c = 8.09; A + C = 161°, A - C = 47° ; £(4 + C) = 80°30', £(^L-C) = 23° 30'. tan 1 (A -C) = ^ tan $(A + C), or, log tan £ (A — C) = log (a — c) + log tan \ (A + C) — log (a + c) log(a-c)= 0.9079 log tan $(A + C) = 10.7764-10 11.6843 - 10 log(a + c) = 2.0458 log tan \ (A - C) = 9. 6386 - 10 .-. £(4-C) = 23°31', which substantially agrees with the above results. « sin A = sin 104° = sin (180° - 104°) = sin 76° . Hence log sin 104° = log sin 76" = 9.9869 - W. 160 PLANE TBIGONOMETKY EXAMPLES Solve the following oblique triangles, using logarithmic Tables I and II. No. Given Parts Required Parts 1 a=10 4 =38° .5=77° 10' C=64°60' 6=15.837 c=14.703 2 a =795 4 = 79° 59' 5=44°41' C=55°20' 6=567.6 c=664 3 6 =.8037 5 =52° 20' C=101°40' 4 = 26° a =.445 c=.9942 4 c = .032 4 = 36° 8' B=U°21' C=99°25' a = . 01913 6=. 02272 5 6=29.01 4 = 87°40' C=33°15' 6 =59° 5' a=33.78 c=18.54 6 a =804 4 = 99° 55' 5=45°1' C=35°4' 6=577.3 c = 468.9 7 a=400 4 = 54° 28' C=60° .5=65° 32' 6=447.4 c=425.7 8 c=161 A = 35° 15' C=123°39' 5=21°6' a=111.6 6=69.62 9 a=5.42 5=42°17.3' C=82°28.4' A = 55° 14.3' 6=4.439 c = 6.542 10 6=2056 4 = 63°52.8' 5=70° C=46°7.2' a =1964.7 c= 1577.3 11 a=7.86 5=32°2'62" C=43°25'26" 4 = 104°31'42" 6=4.309 c=5.583 12 5=8 4=80° 5=2°15'46" C= 97° 44' 14" a =199. 53 c=200.73 Solve the following oblique triangles, using logarithmic Tables I and III. No. Given Parts Eeq uired Parts 13 a = 500 A = 10.2° 6 = 46.6° C = 123.2° 6 = 2051 c = 2363 14 a = 45 .4 = 36.8° C = 62° 5 = 81.2° 6 = 74.25 c = 66.33 15 6 = .085 5 = 95.6° C = 24.2° A = 60.2° a = .0741 c = .035 16 6 = 5685 5 = 48.63° C = 83.26° 4 = 48.11° a = 5640 c = 7523 17 c = 7 4 = 59.58° C = 60° 5 = 60.42° a = 6.971 6 = 7.03 18 c = .0059 5 = 75° C = 36.87° 4 = 68.13° a = .00913 6 = .0095 19 a = 76.08 5 = 126° C = 12.44° 4 = 41.56° 6 = 92.8 c = 24.7 20 a = 22 4 = 3.486° 5 = 73° C= 103.514° 6 = 346 c = 351.8 21 6 = 8000 4 = 24.5° B = 86.495° C = 69.005° a = 3324 c = 7483 22 6 = 129.38 4 = 19.42° C = 64° 5 = 96.58° a = 43.29 c = 117.05 23 c = 95 4 = 2.086° 5 = 112° C = 65.914° a = 3.788 6 = 96.5 24 5 = 132.6 4 = 1° C = 75° 5 = 104° a = 2.385 c = 131.98 25. A ship S can be seen from each of two points 4 and 5 on the shore. By measurement 45 = 800 ft., angle SAB = 67° 43', and angle S54 = 74° 21'. Find the distance of the ship from 4. 4ns. 1253 ft. 26. Two observers 5 mi. apart on a plain, and facing each other, find that the angles of elevation of a balloon in the same vertical plane with them- selves are 65° and 58° respectively. Find the distances of the balloon from the observers. 4ns. 4.607 mi.; 4.45 mi. 27. One diagonal of a parallelogram is 11.237, and it makes the angles 19° 1' and 42° 54' with the sides. Find the sides. 4ns. 4.15 and 8.67. 28. To determine the distance of a hostile fort 4 from a place 5, a line BC and the angles 45C and 5C4 were measured and found to be 322.6 yd., 60° 34', 56° 10' respectively. Find the distance 45. 4ns. 300 yd. THEORY AND USE OF LOGARITHMS 161 29. From points A and B at the bow and stern of a ship respectively, the foremast, C, of another ship is observed. The points A and B are 300 ft. apart, and the angles ABC and BAG are found to be 65.46° and 112.85° respec- tively. What is the distance between the points A and G of the two ships ? Ans. 9254 ft. 30. A lighthouse was observed from a ship to bear N. 34° E. ; after the ship sailed due south 3 mi. it bore N. 23° E. Eind the distance from the lighthouse to the ship in each position. Ans. 6.143 mi. and 8.792 mi. 31. In a trapezoid the parallel sides are 15 and 7, and the angles one of them makes with the nonparallel sides are 70° and 40°. Eind the nonparallel sides. Ans. 8 and 5.47. Case II. When two sides and the angle opposite one of them are given, as a, b, A (ambiguous case *) . First step. Using the law of sines as in Case J, calculate log sinB. If log sinB = 0, sinB = 1, B = 90° j it is a right triangle. If log sinB > 0, sinB > 1 (impossible); there is no solution. If log sinB< and b < a, only the acute value of B found from the table can be used ; there is one solution.^ If log sinB < O and 6 > a, the acute value of B found from the table and also its supplement, should be used ; and there are two solutions, t Second step. Find C (one or two values according as we have one or two values of B) from C = 180°-(A+B). Third step. Find c (one or two values), using law of sines. Check : Use law of tangents. Ex. 1. Having given a = 36, 6 = 80, A = 28° ; solve the triangle. Solution. In attempting to draw a figure of the triangle, the construction appears impossible. To verify this, let us find log sin B in order to apply our tests. First step. Solving -. — r = ^-^= for sinB, * Bini sin_B . _ ftsin^l sm B = , or, log sin B = log 6 + log sin A — log a. log 6= 1.9031 log sin A = 9.6716 - 10 11.5747 - 10 logq= 1.5563 log sin B = 10.0184 -10 = 0.0184. Since log sinB > 0, sinB > 1 (which is impossible), and there is no solution. * In this connection the student should read over 5 58, pp. 104, 105. t For if b<a, B must be less than A, and hence B must be acute. t Since 6>o, A must be acute, and hence B may be either acute or obtuse. 162 PLANE TKlGOJSfOMETRY Ex. 2. Having given o = 7.42, b = 3.39, A = 106°; solve the triangle. Solution. Draw figure. First step. From law of sines, . _ osin.4 sm B = > a log sin B = log 6 + log sin A — log a. log 6= 0.5302 log sin A = 9.9849 - 10 * 10.5151 - 10 logq= 0.8704 log sin B= 9.6447-10 .-. B = 26°11'. Using Table II Since log sin B < and 6 < a, there is only one solution. Second step. O = 180° - (A + B) = 180° - 131° 11' = 48° 49'. Third step. By law of sines, a sin C sini or, log c = log a + log sin C — log sin .A. loga= 0.8704 log sin C = 9.8766 - 10 10.7470 - 10 log sin 4 = 9.9849 - 10 logc= 0.7621 .-. c= 5.783. Check : Use law of tangents. tan£(C - B) = ^^tan £( C + B), or, log tan £ (C - B) = log (c - 6) + log tan £ (C + B) - log (c + 6). Substituting, we find that this equation is satisfied. Ex. 3. Given a = 732, 6 = 1015, A = 40° ; solve the triangle. Solution. It appears from the construction of the triangle that there are two solutions. First step. By law of sines, . _ 6 sin A BinB = , a or, log sin B = log 6 + log sin A — logo. log 6= 3.0065 log sin A = 9.8081-10 = 12.8146 - 10 logq= 2.8645 log sin B= 9.9501-10 * Sin A = sin 105° = sin ( 180° - 105°) = sin 75° . Hence log sin A = log sin 75° = 9.9849 - 10. THEORY AND USE OF LOGARITHMS 163 Since log BinB < and 6 > a, we have two solutions, which test verifies our construction. From Table II we find the first value of B to be B 1 = 6Z°3'. Hence the second value of B is B 2 = 180° - B 1 = 116° 57'. Second step. C x = 180° - {A + B t ) = 180° - 103° 3" = 76° 57' ; C 2 = 180° - (A + B 2 ) = 180° - 156° 57' = 23° 3'. Third step. From law of sines, Cl = o sin Ci or, sin A log c x = log a + log sin Ci — log sin .4. loga= 2.8645 log sin d = 9.9886 - 10 12.8531 - 10 log nmA = 9.8081 - 10 logc 1= 3.0450 In the same manner, from we get .-. Ci = c 2 = 1109.3. a sin C 2 sin A C2 = 445.9. Cheek: Use tan£(C - B) = -tan£(C + B) for both solutions. EXAMPLES Solve the following oblique triangles, using logarithmic Tables I and IL No. Given Pakts Required Parts 1 a=50 c=66 .4 = 123° 11' Impossible 2 a=5.08 6=3.59 A = 63° 50^ _B=39°21' C=76°49 / c=5.511 3 a=62.2 6=74.8 A =27° 18' ^ 1 =33°28 / B 2 =146°32' C 1 = 119°14' C 2 =6°10 / c 1= 118.32 02=14.567 4 6=.2337 c=.1982 5=109° A = 17° 41' C = 53°19' a =.07508 5 a=107 c=171 C=31°53' A = 19° 18' B=128°49 / 6=252.2 6 6=3069 c=1223 C=55°52' Impossible 7 6=6.161 c=6.84 B=44°3' 4 1 =68°47 / 4 2 =23°7 / Ci = 67°10 / C 2 =112°50' ai=6.92 02=2.913 8 a=8.656 c=10 A = 59° 57' JB=30°3' C=90° 6=5.009 9 a=214.56 6=284.79 _B=104°20' 4=46° 53' C=28°47' c=141.5 10 a=32.16 c =27.08 C'=62°24' 4i = 70°12' 4 2 =109°48' Bi=57°24' .B 2 =17°48 / 6 1= 28.79 62 = 10.45 11 6=811.3 c= 606.4 £=126° 6' 20" A = 16° 44' 40" C=37°10' a=289.2 164 PLANE TRIGONOMETRY Solve the following oblique triangles, using logarithmic Tables I and III. No. Given Pakts Required Pakts 12 a = 840 6 = 485 -4 = 21.6° 5 = 12.21° C = 146.29° c = 1272 13 a = 72.63 5 = 117.48 4 = 80° Impossible 14 = 177 6 = 216 A = 35.6° 5 t = 45.27° 5 2 = 134.73° C 1 = 99.13° C 2 = 9.67° ci = 300.3 c 2 = 51.09 15 6 = 9.399 c = 9.197 5 = 120.4° A = 2.02° C= 57.58° a = .3841 16 6 = .048 c = .0621 5 = 57.62° Impossible 17 5 = 19 c = 18 C=15.8° A x = 147.5° 4 2 = 0.9° 5i = 16.7° 5 2 = 163.3° a x = 35.52 a 2 = 1.0385 18 a = 55.55 c = 66.66 C = 77.7° .4 = 54.5° 5 = 47.8° 6=50.54 19 o = 34 c = 22 = 30.35° 4i = 51.37° 4 2 = 128.63° 5 1 = 98.28° 5 2 = 21.02° &i = 43.07 6 2 = 15.613 20 a = 528 6 = 252 4 = 124.6° 5 = 23.14° C = 32.26° c = 342.3 21 6 = 91.06 c = 77.04 5 = 51.12° 4 = 87.69° = 41.19° a = 116.88 22 a = 17,060 6 = 14,050 5 = 40° A r = 51.32° Ci = 88.68° Ci = 21,850 4 2 = 128.68° C 3 = 11.32° c 2 = 4290 23. One side of a parallelogram is 35, a diagonal is 63, and the angle between the diagonals is 21° 37'. Find the other diagonal. Ans. 124.62. 24. The distance from 5 to C is 145 ft. , from A to C is 178 ft., and the angle ABC is 41° 10'. Find the distance from A to 5. Ans. 259.4 ft. 25. Two buoys are 2789 ft. apart, and a boat is 4325 ft. from the nearer buoy. The angle between the lines from the buoys to the boat is 16° 13'. How far is the boat from the further buoy ? Ans. 6667 ft. Case III. When two sides and the included angle are given, as a, b, C* First step. Calculate a+b, a—b; also \{A+B)from A+B=180°— c. Second step. From law of tangents, tani(4-5)=^|tani(,l+5) ; we find ^(A—B~). Adding this result to \(A + B) gives A, and sub- tracting it gives B. Third step. To find side c use law of sines ; for instance, a sin C sin 4 Chech: Chech by law of sines, \ that is, see if log a — log sinA= log b — log sinB = log c — log sin C. * In case any other two sides and included angle are given, simply change the cyclic order of the letters throughout. Thus, if b, c, A are given, use tan - (B - O = ^^ tan - (B + C) , etc. 2 b + c 2 V t From law of sines, a b c s4n A sin B sin C THEORY AND USE OF LOGARITHMS 165 Ex. 1. Having given a = 540, 6 = 420, C = 52° 6' ; solve the triangle, using logarithms from Tables I and II. Solution. Drawing a figure of the triangle on which we indicate the known and unknown parts, we see that the problem comes under Case II, since two sides and the included angle are given. First step. a= 540 540 180° 6 = 420 420 C= 52° 6' a + 6 = 960 a - 6 = 120 A + B = 127° 54' . ^(A + B)= 63° 57'. Second step. tan -(A — 2 V B) = a_6 tan 1 M + J ' + 6 2 V or, Adding, Third step. log tan %(A-B) = log (o - 6) + log tan %(A + B)- log (a + 6). log (a -6)= 2.0792 log tan $(A + B)= 10.3108 - 10 12.3900 - 10 log(o + 6)= 2.9823 log tan i (A-B)= 9.4077 - 10 .-. %(A-B) = U°2V \(A + B) = 63° 57' i(A-B) = U°2\- A = 78° 18'. a sin C sin .A 63° 57' 14° 2 1' Subtracting, B=49°36'. From sin G sin A log c = log a + log sin C — log sin A. log loga = sinC = 2.7324 9.8971 - 10 12.6295-10 log sin A = logc = .-. c = 9.9909-10 2.6386 435.1. Check : By law of sines, log a: log sin A ■■ = 12.7324-10 = 9.9909-10 log 6, = 12.6232 - log sin B= 9.8817 - 2.7415 10 10 2.7415 Ex. 2. Having given a = 167, c arithms from Tables I and III. = 82, B = 98° ; s Solution. First step. a = 167 c= 82 o + c = 249 a 167 82 -c= 85 logc = 12.6386 -10 log sin C = 9.8971 -10 2.7415 solve the triangle, using log- 188° A + C= 82° .-. ^(A+C)= 41°. > 166 PLANE TRIGONOMETRY Second step. ta.n-(A-C)= a — -tani(4 + C), a + c 1 log tan £(4 - C) = log(a - c) + logtan£(4 + C) ■ log(a-c)= 1.9294 logtan£(4 + C) = 9.9392-10 log(a + c) log(a + c): 11.8686 ■ 2.3962 10 Adding, Third step. logtan£(4-C) = 9.4724- -10 .-. £(4-C) = 16.53° £(4 + C) = 41.00° ^(4-C) = 16.53° 4 = 57.53°. 41.00° 16.53° Subtracting, C = 24.47°. , osinJS . b a sin A from sin 2? sin A b = ? log 6 = log a + log sin B — log sin A. logo = 2.2227 log sin B = 9.9958-10 * 12.2185 - 10 log sin A = 9.9262-10 log 6= 2.2923 .-. 6 = 196. Check : By law of sines, logos = 12.2227 -10 log sin A = 9.9262 -10 2.2965 which substantially agree. log 6 = 12.2923 -10 log sin B = 9.9958 - 10 2.2965 logc = 11.9138 -10 log sin C = 9.6172 - 10 2.2966 EXAMPLES Solve the following oblique triangles, using logarithmic Tables I and II. No. Given Parts Required Parts 1 o=27 c = 15 5=46° A = 100° 57' C=33°3' 6=19.78 2 0=486 6=347 C=51°36' 4 = 83° 16' B=45°9' c=383.5 3 6=2.302 e = 3.567 4=62° B=39°16' C = 78°44' 0=3.211 4 0=77.99 6=83.39 C=72°16' 4 = 51° 14.5' 5=56° 29.5' c = 95.24 5 o=0.917 6=0.312 0=33° 7.2' 4 = 132° 18.4 ' B= 14° 33.4' c=.6775 6 a=.3 6=.363 C=124°56' A = 24° 41.8' 5=30° 22.2' c=.5886 7 6=1192.1 c=356.3 A =26° 16' U=143°29' C=10°15' a=886.6 8 a=7.4 c = 11.439 J3=82°26' A =35° 2' C=62°32' 6=12.777 9 a=53.27 6=41.61 C=78°33' 4 = 59° 16.5' .6=42° 10.5' c = 60.74 10 6 =.02668 c = . 05092 A = 115° 47' B=21°1.2' C=43°11.8' o=. 06699 11 a = 51.38 c=67.94 B= 79° 12' 36" 4 = 40° 52.7' C= 59° 64.7' 6=77.12 12 5=V3 c = V3 A =35° 53' B= 93° 28.5' C= 50° 38.5' a=1.313 * sin B= sin 98°= sin (180° -98°)= sin 82°. .-. log sin 98°= log sin 82° =9.9958- 10. THEORY AND USE OF LOGARITHMS 167 Solve the following oblique triangles, using logarithmic Tables I and III. No. Given Parts Required Parts 13 a = 17 6= 12 C = 59.3° A = 77.2° £ = 43.5° c = 14.99 14 a = 55.14 6 = 33.09 C = 30.4° A = 117.4° B = 32.2° c = 31.43 15 6 = 101 c = 158 A = 37.38° B = 38.26° C = 104.36° a = 99.04 16 a = 101 6 = 29 C = 32.18° A = 136.4° B = 11.42° c= 78 17 c = 45 6 = 29 A = 42.8° B = 39.72° C = 97.48° a = 30.84 18 a = .085 c = .0042 B = 56.5° A = 121.07° C = 2.43° 6 = .08276 19 6 = .9486 c = .8852 A = 84.6° B = 49.88° C = 45.52° a = 1.235 20 6 = 6 c = 9 A = 88.9° B = 34.03° C= 57.07° a= 10.72 21 a = 12 6 = 19 C = 5.24° A = 8.84° B = 165.92° c = 7.132 22 a = 42,930 c = 73,480 .5 = 24.8° X = 27.56° C= 127.64° 6 = 38,920 23. In order to find the distance between two objects, A and B, separated by a swamp, a station C was chosen, and the distances CA = 3825 yd., CB = 3476 yd., together with the angle -4CB = 62° 31', were measured. "What is the distance AB ? Ans. 3800 yd. 24. Two trains start at the same time from the same station and move along straight tracks that form an angle of 30°, one train at the rate of 30 mi. an hour, the other at the rate of 40 mi. an hour. How far apart are the trains at the end of half an hour ? Ans. 10.27 mi. 25. In a parallelogram the two diagonals are 5 and 6 and form an angle of 49° 18'. Find the sides. Ans. 5.004 and 2.339. 26. Two trees A and B are on opposite sides of a pond. The distance of A from a point C is 297.6 ft., the distance of B from C is 864.4 ft., and the angle ACB is 87.72°. Find the distance AB. Ans. 903 ft. 27. Two stations A and B on opposite sides of a mountain are both visible from a third station C. The distances AC, BC, and the angle ACB were meas- ured and found to be 11.5 mi., 9.4 mi., and 59° 31' respectively. Find the dis- tance from A to B. Ans. 10.535 mi. 28. From a point 3 mi. from one end of an island and 7 mi. from the other end the island subtends an angle of 33° 55.8'. Find the length of the island. Ans. 4.814 mi. 29. The sides of a parallelogram are 172.43 and 101.31, and the angle included by them is 61° 16'. Find the two diagonals. Ans. 152.33 and 238.3. 30. Two yachts start at the same time from the same point, and sail, one due north at the rate of 10.44 mi. an hour, and the other due northeast at the rate of 7.71 mi. an hour. How far apart are they at the end of 40 minutes ? Ans. 4.927 mi. Case IV. When all three sides a, b, c are given. First step. Calculate s = ^ (a + b + e), s — a, s — b, s — c. Second step. Find log r from , = ^ (s-")(s-l>)(s-°) . (84 ) to (87), p. 115 168 PLANE TRIGONOMETRY Third step. Find angles A, B, C from r tan \A tan ^ B = ■ r r — - > tan A C = — — b 2 — o s — a Check : SeeifA + B + C= 180°. Ex. 1. Having given a = 51, 6 = 65, c = 20 ; solve the triangle. Solution. Drawing a figure of the triangle on which we indicate the known and unknown parts, we see that since the three sides are given, the problem comes under Case IV. First step, a = 51 6= 65 c= 20 2 s = 136 Hence Second step. s = 68 a= 51 a = 17 s = 68 6 = 65 s-6= 3 s = 68 c = 20 s - c = 48 j(s - a) (s -b)(s-e) or, logr = £[log(s - a) + log(s - 6) + log(s - c) - logs]. From the table of logarithms, 6 = 65 Third step. From the formula tan £ A log(s- a) = 1.2304 log (s- 6) = 0.4771 log (s-c) = 1.6812 3.3887 logs =1.8325 2 1 1.5562 logr = 0.7781 s — a log tan ^ A = log r — log (s — a). logr = 10.7781 -10 log(s-q)= 1.2304 or, From the formula tan £ B = \A= 19.44° and .4=38.88° Check : A + B+C= 179.98". log tan £4= 9.5477-10 using Table II* £J. = 19°27', A - 38° 54'. s — 6 log tan £ _B = logr — log(s -6). logr = 10.7781 -10 log (s -6)= 0.4771 log tan £ B = 10.3010 - 10 £.8 = 63° 26', using Table II B = 126° 52'. ad, we get , $B= 63.43°, 5 C= 7.12°, , J3 = 126.86°, C= 14.24°. THEORY AND USE OF LOGARITHMS 169 From the formula tan £ O = ■ Check : log tan I G = log r — log (s — c). logr = 10.7781 -10 log (s-c)= 1.6812 log tan £ C = 9.0969 - 10 ^ C = 7° 8'. C = 14° 16'. A = 38° 54' £ = 126° 52' C = 14° 16' A + B + C = 180° 2' Using Table II EXAMPLES Solve the following oblique triangles, using logarithmic Tables I and II. No. Given Parts Required Parts 1 a = 2 6 = 3 c = 4 .4 = 28° 58' £ = 46° 34' C = 104° 28' 2 a = 2.5 6 = 2.79 c = 2.33 A = 57° 38' B = 70° 28' O = 51° 54' 3 a = 5.6 6 = 4.3 c = 4.9 ^4 = 74° 40' B = 47° 46' C = 57° 34' 4 a= 111 6 = 145 c = 40 A = 27° 20' B = 143° 8' C = 9° 32' 5 a = 79.3 6 = 94.2 c = 66.9 .4. = 65° 56' B = 79° 44' C = 44° 20' 6 a= 321 6 = 361 c = 402 A = 49° 24' B = 58° 38' C = 71° 58' 7 a= .641 b = .529 c = .702 A = 60° 52' £ = 46° 6' C = 73°2' 8 a = 3.019 6 = 6.731 c = 4.228 A = 18° 12' £ = 135° 52' C = 25° 56' 9 a= .8706 6 = .0916 c = .7902 A = 149° 50' £ = 3° 2' C = 27° 10' 10 a = 73 6 = 82 c = 91 4 = 49° 34' £ = 58° 46' C = 71° 38' 11 a = 1.9 6 = 3.4 c = 4.9 A = 16° 26' £ = 30° 24' C = 133° 10' 12 o= .21 6 = .26 c = .31 A = 42° 6' £ = 56° 6' C = 81°48' 13 a = 513.4 6 = 726.8 c = 931.3 A = 33° 16' £ = 50° 56' C = 95°48' 14 a=V5 6 = V6 c=V7 X = 51° 52' £= 59° 32' C = 68° 34' Solve the following oblique triangles, using logarithmic Tables I and III. No. Given Parts Required Parts 15 a = 4 6 = 7 c = 6 ^. = 34.78° £ = 86.42° C= 58.82° 16 a = 43 6 = 50 c = 57 A = 46.82° £ = 57.98° C = 75.18° 17 a= .23 6= .26 c = .198 A = 58.44° £ = 74.38° C= 47.18° 18 a = 61.3 6 = 84.7 c = 47.6 A = 45.2° £ = 101.38° C = 33.44° 19 a = .0291 6= .0184 c = .0358 ^.= 54.06° £ = 30.8° C = 95.16° 20 a= 705 6 = 562 c = 639 A = 71.56° £ = 49.14° C = 59.32° 21 a = 56 6 = 43 c = 49 ^. = 74.68° £ = 47.78° C = 57.56° 22 a = 301.9 6 = 673.1 c = 422.8 A = 18.2° £= 135.86° C = 25.94° 23 a = 2.51 6 = 2.79 c = 2.33 A = 57.88° £ = 70.3° C= 51.84° 24 a = 80 6=90 c = 100 ^ = 49.46° £ = 58.76° G = 71.78° 170 PLANE TRIGONOMETRY 25. The sides of a triangular field are 7 rd., 11 rd., and 9.6 rd. Find the angle opposite the longest side. Ans. 81° 22'. 26. A pole 13 ft. long is placed 6 ft. from the base of an embankment, and reaches 8 ft. up its face. Find the slope of the embankment. Ans. 44° 2'. 27. Under what visual angle is an object 7 ft. long seen when the eye of the observer is 5 ft. from one end of the object and 8 ft. from the other end ? Ans. 60°. 28. The distances between three cities, A, B, and C, are as follows ; AB = 166 mi., AC = 72 mi., and BC = 186 mi. B is due east from A. In what direction is from A ? Ans. N. 4° 24' W. or S. 4° 24' W. 29. Three towns, A, B, and C, are connected by straight roads. AB = 4 mi., BC = 5 mi. , A C = 1 mi. Find the angle made by the roads AB and BC. Ans. 101.55°. 30. The distances of two islands from a buoy are 3 and 4 mi. respectively. If the islands are 2 mi. apart, find the angle subtended by the islands at the buoy. Ans. 28.96°. 31. A point P is 13,581 ft. from one end of a wall 12,342 ft. long, and 10,025 ft. from the other end. What angle does the wall subtend at the point P ? Ans. 60.86°. 83. Use of logarithms in finding the area of an oblique triangle. From § 62, p. 117, we have the following three eases. Case I. When two sides and the included angle are given, use one of the formulas /oox aft sinC be sin A ac sin B (88) S= , S= , S = , 2 2 2 where S = area of the triangle. Ex. 1. Given a = 25.6, 5 = 38.2, C = 41° 56' ; find the area of the triangle. „ , „ a&sinC Solution. S = 2 logS = logo + logo + logsinC - k>g2. loga= 1.4082 log 6= 1.5821 log sin C = 9.8249 - 10 12.8152 - 10 log2 = 0.3010 logS= 12.5142-10 = 2.5142. .-. 8 = 326.8. Ans. Case II. When the three sides are given, use formula (89) S = Vs(s-a)(s-&)(s-c), where S = area of the triangle, and s = ] s (a + b + c). THEORY AND USE OF LOGARITHMS 171 Ex. 2. Find the area of a triangle, having given a =12.53, 6 = 24.9, c =18.91. Solution, a = 12.53 Hence 6 = 24.9 s = 28.17 s = 28.17 s = 28.17 c = 18.91 a = 12.53 6 = 24.9 c = 18.91 2 s = 56.34 s-a = 15.64 s - b = 3.27 s-c= 9.26 s = 28.17. S = Vs (s - a) (s - 6) (s - c). log S = £ [log s + log (s - a) + log (s -6) + log(s- logs = 1.4498 log(s - a) = 1.1942 log (J i-6) = 0.5145 logO i - c) = 0.9666 2 1 4.1251 logS = 2.0626 .-. 8 = 115.5. Ans. «)] Case III. ^irea problems which do not fall directly under Cases I or II may be solved by Case I if we first find an additional side or angle by the law of sines. Ex. 3. Given A = 34° 22', B = 66° 11', c = 78.35 ; find area of triangle. Solution. This does not now come directly under either Case I or Case II. But C = 180° - (A + B) = 180° - 100° 33' = 79° 27'. And, by law of sines, csin.4 smC log a = log c + log sin A — log sin C. logc= 1.8941 log sin 4 = 9.7517 -10 11.6458 - 10 log sin C = 9.9926 - 10 loga= 1.6532 Now it comes under Case I. ac sin B o = ■ 2 logS = loga + logc + logsinJJ - log 2. loga= 1.6532 logc= 1.8941 log sin B = 9.9614-10 13.5087 - 10 log 2 = 0.3010 logS = 13.2077 -10 = 3.2077 .-. S = 1613.3. Ans. 172 PLANE TRIGONOMETRY EXAMPLES Find the areas of the following oblique triangles, using Tables I and II for the first ten and Tables I and III for the rest. No. Given Paets Area i a = 38 c = 61.2 B = 67° 56' 1078 2 6 = 2.07 .4 = 70° B = 36° 23' 3.257 3 6 = 116.1 c= 100 A = 118° 16' 5113 4 a = 3. 123 A = 53° 11' B = 13° 57' 1.354 5 6 = .43,9 A = 76° 38' C = 40° 35' .0686 6 a = .3228 c= .9082 B = 60° 16' .1273 7 c = 80.25 B = 100° 5' C = 31° 44' 4494 8 a = .010168 6= .018225 C = 11° 18.4' .000018155 9 a = 18.063 A = 96° 30' B = 35° 70.55 10 6= 142.8 c = 89.0 a = 95 4174 11 a = 100 B = 60.25° C = 54.5° 3891 12 a = 145 6 = 178 5 = 41.17° 12,383 13 a = 886 6 = 747 C = 71.9° 314,600 14 a =266 6 = 352 C = 73° 44,770 15 a = 960 6 = 720 = 25.67° 149,730 16 a = 79 6 = 94 c = 67 2604 17 a = 23.1 6 = 19.7 c = 25.2 215.9 18 a = 5.82 6 = 6 c = 4.26 11.733 19. The sides of a field ABCD are AB = 37 rd. , BC = 63 rd. , and DA = 20 rd., and the diagonals A C and BD are 75 rd. and 42 rd., respectively. Required the area of the field. 1570 sq. rd. 20. In a field ABCD the sides AB, BC, CD, and DA are 155 rd., 236 rd., 252 rd., and 105 rd., respectively, and the length from A to C is 311 rd. Find the area of the field. • 29,800 sq. rd. 21. The area of a triangle is one acre ; two of its sides are 127 yd. and 150 yd. Find the angle between them. 30° 32'. 22. Given the area of a triangle =12. Find the radius of the inscribed circle if a = 60 and B = 40° 35.2'. 84. Measurement of land areas. The following examples illustrate the nature of the measurements made by surveyors in determining land areas, and the usual method employed for calculating the area from the data found. The Gunter's chain is 4 rd., or 66 ft., in length. An acre equals 10 sq. chains, or 160 sq. rd. EXAMPLES 1. A surveyor starting from a point A runs N. 27° B. 10 chains to B, thence N.E. by E. 8 chains to C, thence S. 5° W. 24 chains to D, thence N. 40° 44' W. 13.94 chains to A. Calculate the area of the field ABCD. Solution. Draw an accurate figure of the field. Through the extreme west- erly point of the field draw a north-and-south line. From the figure, area THEORY AND USE OF LOGARITHMS 173 iM ABCD = area trapezoid* GCDE — (area trapezoid QOBF + area triangle .F25.4 + area triangle ABE) = 13.9 acres. Ans. 2. A surveyor measures S. 50°25'E. 6.04 chains, thence S. 58°10'W. 4.15 chains, thence N. 28° 12' W. 5.1 chains, thence to the starting point. Determine the direction and distance of the starting point from the last station, and find area of the field inclosed. Ans. N. 39° 42' E. 2 chains ; 1.66 acres. 3. One side of a field runs N. 83° 30' W. 10.5 chains, the second side S. 22° 15' W. 11.67 chains, the third side N. 71° 45' E. 12.9 chains, the fourth side completes the circuit of the field. Find the direction and length of the fourth side, and calculate the area of the field. Ans. N. 25° 1' E. 6.15 chains ; 8.78 acres. 4. From station No. 1 to station No. 2 is S. 7° 20' W. 4.57 chains, thence to station No. 3 S. 61° 55' W. 7.06 chains, thence to station No. 4 N. 3° 10' E. 5.06 chains, thence to station No. 5 N. 33° 50' E. 3.25 chains, thence to station No. 1. Find the direction and distance of station No. 1 from No. 5, and calculate the area of the field inclosed. Ans. E. 1° 15' N. 4.7 chains ; 3.55 acres. 85. Parallel sailing. When a vessel sails due east or due west, that is, always travels on the same parallel of latitude, it is called parallel sailing. The distance sailed is the departure,^ and it is expressed in geographical % miles. Thus, in the figure, arc AB is the departure between A and B. The latitudes of A and B are the same, i.e. arc EA = angle EOA = arc QB = angle QOB. The differenqe in longitude » From Geometry the area of a trapezoid equals one half the sum of the parallel sides times the altitude. Thus, area GCDE = \(GC + ED) GE. t The departure between two meridians is the arc of a parallel of latitude comprehended between those meridians. It diminishes as the distance from the equator increases. X A geographical mile or knot is the length of an arc of one minute on a great circle of the earth. 174 PLANE TRIGONOMETRY of A and B = arc EQ. The relation between latitude, departure, and difference in longitude may be found as follows : By Geometry, axoAB DA DA A A „ , ,_., , = = = cos 0^4 -D =t= cos A OE = cos latitude. arcEQ OE OA (90) arc AB = arcEQ cos latitude, or, departure Diff. long. = cos latitude EXAMPLES 1. A ship whose position is lat. 25° 20' N., long, knots. Find the longitude of the place reached. Solution. Here departure = 140, and latitude = 25° 20' N. 6° 10' W. sails due west 140 Substituting in above formula (90), diff. long. = 140 cos 25° 20' log 140 = 12.1461 -10 log cos 25° 20' = 9.9561 - 10 log diff . long. = 2.1900 diff. long. = 154.9' = 2° 34.9'. Hence longitude of place reached = 36° 10' + 2° 34.9' = 38° 44.9' W. Ana. 2. A ship in lat. 42° 16' N., long. 72° 16' W., sails due east a distance of 149 geographical miles. "What is the position of the point reached ? Ans. Long. 68° 55' W. 3. A vessel in lat. 44° 49' S., long. 119° 42' E., sails due west until it reaches long. 117° 16' E. Eind the departure. Ans. 103.6 knots. 4. A ship in lat. 36° 48' N., long. 56° 15' W., sails due east 226 mi. Find the longitude of the place reached. Ans. Long. 51° 33' W. 5. A vessel in lat. 48° 54' N., long. 10° 55' W., sails due west until it is in long. 15° 12' W. Find the number of knots sailed. Ahs. 168.9 knots. 86. Plane sailing. When a ship sails in such a manner as to cross successive meridians at the same angle,, it is said to sail on a rhumb line. This , angle is called the course, and the distance between two places is measured on p rhumb line. Thus, in the figure, if* 5 ^ ship travels from A to B on a rhumb lint. arc AB = distance, angle CAB = course, arc CB = departure, arc A C = difference in latitude between A and B. (.North Pole) p gqOM ,aW THEORY AND USE OF LOGAEITHMS 175 An approximate relation between the quantities involved is ob- tained by regarding the surface of the earth as a, plane surface, that is, regarding A CB as a plane right triangle, the angle A CB being the right angle. This right triangle is called the triangle of plane sailing. From this plane right triangle we get CB = ABsinA, and ' -""'"■' A C = AB cos A ; or, (91) Departure = distance x sin course, and (92) Diff. lat. = distance x cos course. If AB is long, the error caused by neglecting the curvature of the earth will be too great to make these results of any value. In that case AB may be divided into parts, such as AE, EG, GI, IB (figure on p. 174), which are so small that the curvature of the earth may be neglected. EXAMPLES 1. A ship sails from lat. 8° 45' S., on a course N. 36° E. 345 geographical mi. Find the latitude reached and the departure made. Solution. Here distance = 345 and course = 36°. .-. departure = 345 sin 36°. diff. lat. = 345 cos 36°. log345 = 2.5378 log345 = 2.5378 log sin 36° = 9.7692-10 log cos 36° = 9.9080 -10 log departure = 2. 3070 log diff. lat. = 2.4458 .-. departure = 202.8 mi. Ans. diff. lat. = 279. 1' = 4° 39. V. As the ship is sailing in a northerly direction she will have reached latitude go 45 / _ 4 o 39 y _ 4 o 5 9 x s _ Ans 2. A ship sails from lat. 32° 18' N., on a course between N. and W., a dis- tance of 344 mi., and a departure of 103 mi. Find the course and the latitude reached. Ans. Course K 17° 25' W. , lat. 37° 46' N. 3. A ship sails from lat. 43° 45' S., on a course N. by E. 2345 mi. Find the latitude reached and the departure made. Ans. Lat. 5° 25' S., departure = 457.5 mi. 4. A ship sails on a course between S. and E. 244 mi., leaving lat. 2° 52' S., and reaching lat. 6° 8' S. Find the course and the departure. Ans. Course S. 56° 8' E., departure = 202.6 mi. 87. Middle latitude sailing. Here we take the departure between two places to be measured on that parallel of latitude which lies halfway between the parallels of the two places. Thus, in the figure on p. 174, the departure between A and B is LM, measured on a parallel of latitude midway between the parallels of A and B- 176 PLANE TRIGONOMETRY This will be sufficiently accurate for ordinary purposes if the run is not of great length nor too far away from the equator. The mid- dle latitude is then the mean of the latitudes of A and B. The formula (90) on p. 174 will then become departure (93) Diff. long. = cos mid. lat. EXAMPLES 1. A ship in lat. 42° 30' N., long. 58° 51' W., sails S. 33° 45' E. 300 knots. Find the latitude and longitude of the position reached. Solution. We know the latitude of the starting point A. To get the latitude of the final position B, we first find diff. in lat. from (92). This gives diff. lat. = 300 cos 33° 45'. log 300 = 2.4771 log cos 33° 45' = 9.9198 - 10 log diff. lat. = 2.3969 diff. lat. = 249.4' =4° 9.4'. Since the ship sails in a southerly direction, she will ,, , 7 ... havereaohedlatitude=42 30'-4°9.4'=38°20.6'N. Ans. To get the longitude of B we must first calculate the departure and middle latitude for substitution in (93). From (91) departure = 300 sin 33° 45'. log 300= 2.4771 log sin 33° 45' = 9.7448 - 10 log departure = 2.2219 departure = 166.7'. Middle latitude = J (42? 30' + 38° 20.6') = 40° 26.3'. "1 ftfi 7 Substituting in (93), diff . long. = '■ 8 v '' 5 cos 40° 25.3' log 166.7= 12.2219-10 log cos 40° 25. 3' = 9.8815 - 10 log diff. long. = 2.3404 diff. long. = 219' = 3° 39'. Since the ship sails in an easterly direction, she will have reached longitude = 58° 51' - 3° 39' = 65° 12' W. Ana. 2. A vessel in lat. 26°15'N., long. 61°43'W., sails N.W. 253 knots. Find the latitude and longitude of the position reached. Ans. Lat. 29° 13.9' N. ; long. 65° 5.1' W. 3. A ship leaves lat. 31° 14' N., long. 42° 19' W., and sails E.N.E. 325 mi. Find the position reached. Ans. Lat. 33° 18.4' N. ; long. 36° 24' W. 4. Leaving lat. 42° 30' N. , long. 58° 51' W., a battleship sails S. E. by S. 300 mi. Find the place reached. Ans. Lat. 38° 21' N. ; long. 55° 12' W. THEORY AND USE OF LOGARITHMS 177 6. A ship sails from a position lat. 49° 56' N., long. 15° 16' W., to another lat. 47° 18' N., long. 20° 10' W. Find the course and distance. Ans. Course, S. 50° 53' W. ; distance = 250.5 mi. Hint. The difference in latitude and the difference in longitude are known, also the middle latitude. 6. A torpedo boat in lat. 37° N. , long. 32° 16' W. , steams N. 36° 56' W. , and reaches lat. 41° N. Find the distance steamed and the longitude of the position reached. Ans. Distance = 300.3 mi.; long. 36°" 8' W. 7. A ship in lat. 42° 30' N., long. 58° 51' W., sails S.E. until her departure is 163 mi. and her latitude 38° 22' N. Eind her course and distance and the longi- tude of the position reached. Ans. Course, S. 33° 19' E. ; distance = 296.7 mi. ; long. 55° 17' W. 8. A cruiser in lat. 47°44'K, long. 32°44'W., steams 171 mi. N.E. until her latitude is 50° 2' N. Find her course and the longitude of the position reached. Ans. Course, N. 36° 11' E. ; long. 30° 10' W. 9. A vessel in lat. 47° 15' N. , long. 20° 48' W. , sails S. W. 208 mi. , the departure being 162 mi. Find the course and the latitude and longitude of the position reached. Ans. Course, S. 51° 9' W. ; lat. 45° 4.5' N. ; long. 24°42'W. CHAPTEE IX ACUTE ANGLES NEAR 0° OR 90° 88. When the angle x approaches the limit zero, each of the ratios , , approaches unity as a limit, x being the circular measure xx of the angle. Proof. Let be the center of a circle whose radius is unity. Let arc AP = x, and let arc AP' = x in numerical value. Draw PP', and let PT and P'T be the tangents drawn to the circle at P and P'. From Geometry But PQP' < PAP' < PTP'. PQP' = PQ -j- QP' = 2 sin x in numerical value, PAP' = PA + AP' = 2 x in numerical value, and PTP' = PT+ TP' = 2 tan x in numerical value. Substituting in (A). 2 sin a; < 2 a; < 2 tan x. Dividing through by 2, we have (B) sin* < x < tan x, which proves that If x be the circular measure of an acute angle, it will always lie be- tween sin x and tan x, being greater than sin x and less than tan x. Dividing (B) through by sin x, we get x 1 K < sin x cos x 11 we now let x approach the limit zero, it is seen that x limit. x=0 sin a; must lie between the constant 1 and " ml *' , x =° cos a; 178 which is also 1. ACUTE ANGLES NEAR 0° OE 90° 179 Hence limit -?— = 1, or, SIM (C) limit f^f = 1 Similarly, if we divide (B) through by tan x, we get x cos x < - < 1. tana; As before, if x approaches zero as a limit, limit tana: a:= g; must lie between the constant 1 and hmit cos x w hich is also 1. x=0 ' . Hence limit -^- = 1, or, x =° tana; ' (D) limit ^1 = 1 The limits (C) and (D) are of great importance both in pure and applied mathematics. These results may be stated as follows : When x is the circular measure of a very small angle we may replace sin x and tan x in our calculations by x. 89. Functions of positive acute angles near 0° and 90°. So far we have assumed that the differences in the trigonometric functions are proportional to the differences in the corresponding angles. While .this is not strictly true, it is in general sufficiently exact for most practical purposes unless the angles are very near 0° or 90°. In using logarithms we have also assumed that the differences in the logarithms of the trigonometric functions are proportional to the differences in the corresponding angles. This will give sufficiently accurate results for most purposes if we use Tables II or III in the tables and confine ourselves to angles between .3° (= 18') and 89.7° (= 89° 42') inclusive. If, however, we have an angle between 0° and .3°(= 18') or one between 89.7° (= 89° 42') and 90°, and are looking for exact results, it is evident that the ordinary method will not do. For example, the tabular difference (Table II) between the logarithmic sine, tangent, or cotangent of 8' and the logarithm of the corresponding functions of 9' is 512, while between 9' and 10' it is 457. If we interpolate here in the usual way it is evident that 180 PLANE TRIGONOMETRY our results will be only approximately correct. In case it is desired to obtain more accurate results we may use the principle established in the last section, namely : We may replace sin x and tan x in our calculations by x when x is a very small angle and is expressed in circular measure. Prom a table giving the natural functions of angles, we have sin 2.2°= 0.03839 = 0.0384, tan 2.2°= 0.03842 = 0.0384. Also 2.2° = 0.0384 radians. Hence it is seen that in any calculation we may replace the sine or tangent of any angle between 0° and 2.2° by the circular measure of the angle without changing the first four significant figures of the result. Also since cos 87.8°= sin (90°- 87.8°) = sin 2.2°= 0.0384, cot 87.8° = tan (90° - 87.8°) = tan 2.2° = 0.0384, and 2.2° = 90° - 87.8° = 0.0384 radians, we may replace the cosine or cotangent of any angle between 87.8° and 90° by the circular measure of the complement of that angle. We may then state the following rules : 90. Rule for finding the functions of acute angles near 0°. sin x = circular measure of x,* tanx = circular measure of x, cot x = — ; — ! circular measure oj x cos x is found from the tables in the usual way.^ * The following equivalents may be used for reducing an angle to circular measure (radians), and in other computations. 1° = Ton radians. 1° = 0.0174533 radians, log 0.0174533= 8.2419 - 10. 1' = 0.0002909 radians, log0.0002909= 6.4637 - 10. 1" = 0.0000048 radians, log0.0000048 = 4.6856- 10. 180° =^-=57.29578° = 1 radian, log57.29578= 1.7581. T =3.14159 log tt= 0.4971. = %p approximately. t esc a: and sec a: are simply the reciprocals of sinx and cos a; respectively. ACUTE ANGLES NEAR. 0° OB, 90° 181 91. Rule for finding the functions of acute angles near 90°. cos x = circular measure of the complement of x,* cot x = circular measure of the complement of x, 1 circular measure of the complement of x sin x is found from the tables in the usual way.f Since any function of an angle of any magnitude whatever, posi- tive or negative, equals some function of a positive acute angle, it is evident that the above rules, together with those on p. 57, will suffice for finding the functions of angles near ±90°, ±180°, ±270°, ±360°. Ex. 1. Find sine, tangent, and cotangent of 42'. Solution. Reducing the angle to radians, 42' = 42 x 0.0002909 radians = 0.01222 radians. Therefore sin 42' = 0.01222, tan 42' = 0.01222, cot 42' = — - — = 81.833. Ans. 0.01222 Ex. 2. Find cosine, cotangent, and tangent of 89° 34.6'. Solution. The complement of our angle is 90° - 89° 34.6' = 25.4'. Reducing this remainder to radians, 25.4' = 25.4 x 0.0002909 radians = 0.00739 radians. Therefore cos 89° 34. 6' = 0. 00739, cot 89° 34. 6' =0.00739, tan 89° 34. 6' = — - — = 135. 32. Ans. 0.00739 When the function of a positive acute angle near 0° or 90° is given, to find the angle itself we reverse the process illustrated above. For instance: Ex. 3. Eind the angle subtended by a man 6 ft. tall at a distance of 1225 ft. Solution. _ lI ■ | 6 From the figure tan x = T -$$-g. " 122 6 But, since the angle is very small, we may replace tan x by x, giving x — j^-g radians = 0.0049 radians. Or, reducing the angle to minutes of arc, we get 0.0049 „ -la a, a — z — minutes of arc = 16.8 . Ans. 0.0002909 * If the angle is given in degrees, subtract it from 90° and reduce the remainder to cir. cular measure (radians). If the angle is given in circular measure (radians), simply subtract it from - (= 1.57079). 2 t esc x and sec a; are simply the reciprocals of sin x and cos x respectively. 182 PLANE TRIGONOMETRY 92. Rules for finding the logarithms of the functions of angles near 0° and 90°.* For use in logarithmic computations trie rules of the last two sections may be put in the following form : If the angle is given in degrees, minutes, and seconds, it should first be reduced to degrees and the decimal part of a degree (see Conver- sion Table on p. 17 of Tables'). Rule I. To find the logarithms of the functions of angles near 0°. log sin x° = 2.2419 + log re.f log tan x" = 2.2419 + log x. log cot x° = 1.7581 — log x.% log cos x" is found from the tables in the usual way. Rule II. To find the logarithms of the functions of an angle near 90°. log cos x° = 2.2419 + log (90 — x). log cot x" = 2.2419 + log (90 — x). log tanx" = 1.7581 — log (90 — x). log sin x° is found from the tables in the usual way. Ex. 1. Find log tan 0.045°. Solution. As is indicated in our logarithmic tables, ordinary interpolation will not give accurate results in this case. But from the above rule, log tan 0.045° = 2.2419 + log 0.045 = 2.2419 + 2.6532. .-. log tan 0.045° =4.8951. Ans. On consulting a much larger table of logarithms, this result is found to be exact to four decimal places. Interpolating in the ordinary way, we get log tan 0.045° = 4.8924, which is correct to only two decimal places. * These rules will give results accurate to four decimal places for all angles between 0° and 1.1° and between 88.9° and 90°. t Since 1 degree = 0.017453 radians, the circular measure of x degrees = 0.017453 • x radians. Hence, from p. 180, sin x° = 0.017453 ■ x, and log sin x° = log 0.017453 + log x = 2.2419+ log x. % From p. 180, cot x° = , C 017453 • x and log cot x" = - log 0.017453 — log x = 1.7581 -log a;. ACUTE ANGLES NEAR 0° OR 90° 183 Ex. 2. Find log tan 89.935°. Solution. From the above rule, log tan 89.935° = 1.7581 - log (90 - 89.935) = 1.7581 -log 0.065 = 1.7581-2.8129. .-. log tan 89.935° =2.9452. Ans. If the tangent itself is desired, we look up the number in Table I correspond- ing to this logarithm. This gives tan 89.935° =881.4. 93. Consistent measurements and calculations. In the examples given so far in this book it has generally been assumed that the given data were exact. That is, if two sides and the included angle of a triangle were given, as 135 ft., 217 ft., and 25.3° respectively, we have taken for granted that these numbers were not subject to errors made in measurement. This is in accordance with the plan followed in the problems that the student has solved in Arithmetic, Algebra, and Geometry. It should not be forgotten, however, that when we apply the principles of Trigonometry to the solution of practical prob- lems, — engineering problems, for instance, — it is usually necessary to use data which have been found by actual measurement, and there- fore are subject to error. In taking these measurements one should carefully see that they are made with about the same degree of accuracy. Thus, it would evidently be folly to measure one side of a triangle with much greater care than another, for, in combining these measurements in a calculation, the result would at best be no more accurate than the worst measurement. Similarly, the angles of a triangle should be measured with the same care as the sides. The number of significant figures in a measurement is supposed to indicate the care that was intended when the measurement was made, and any two measurements showing the same number of sig- nificant figures will, in general, show about the same relative care in measurement. If the sides of a rectangle are about 936 ft. and 8 ft., the short side should be measured to at least two decimal places. A neglected 4 in the tenths place will alter the area by 374 sq. ft. The following directions will help us to make consistent measure- ments and avoid unnecessary work in our calculations. 1. Let all measured lines and calculated lines show the same num- ber of significant figures, as a rule. 2. When the lines show only one significant figure, let the angles read to the nearest 5°. 184 PLANE TRIGONOMETRY 3. When the lines show two significant figures, let the angles read to the nearest half degree. 4. When the lines show three significant figures, let the angles read to the nearest 5'. 5. When the lines show four significant figures, let the angles read to the nearest minute. EXAMPLES 1. The inclination of a railway to the horizontal is 40'. How many feet does it rise in a mile 1 Arts. 61.43. 2. Given that the moon's distance from the earth is 238,885 mi. and subtends an angle of 31' 8" at the earth. Find the diameter of the moon in miles. Ana. 2163.5. 3. Given that the sun's distance from the earth is 92,000,000 mi. and subtends an angle of 32' 4" at the earth. Find the sun's diameter. Ans. 858,200 mi. 4. Given that the earth's radius is 3963 mi. and subtends an angle of 57' 2" at the moon. Find the distance of the moon from the earth. Ans. 238,833 mi. 5. Given that the radius of the earth is 3963 mi. and subtends an angle of 9" at the sun. Find the distance of the sun from the earth. Ans. 90,840,000 mi. 6. Assuming that the sun subtends an angle of 32' 4" at the earth, how far from the eye must a dime be held so as to just hide the sun, the diameter of a dime being | in. 1 Ans. 76.6 in. 7. Find the angle subtended by a circular target 5 ft. in diameter at the distance of half a mile. Ans. 6' 30.6". MISCELLANEOUS EXAMPLES 1. A balloon is at a height of 2500 ft. above a plain and its angle of elevation at a point in the plain is 40° 35'. How far is this point from the balloon ? Ans. 3843 ft. 2. A tower standing on a horizontal plain subtends an angle of 37° 19.5' at a point in the plain distant 369.5 ft. from the foot of the tower. Find the height of the tower. Ans. 281.8 ft. 3. The shadow of a steeple on a horizontal plain is observed to be 176.23 ft. when the elevation of the sun is 33.2°. Find the height of the tower. Ans. 115.3 ft. 4. From the top of a lighthouse 112.5 ft. high, the angles of depression of two ships, when the line joining the ships passes through the foot of the light- house, are 27.3° and 20.6° respectively. Find the distance between the ships. Ans. 81 ft. 5. From the top of a cliff the angles of depression of the top and bottom of a lighthouse 97.25 ft. high are observed to be 23° 17' and 24° 19' respectively. How much higher is the cliff than the lighthouse ? Ans. 1947 ft. 6. The angle of elevation of a balloon from a station due south of it is 47° 18.5', and from another station due west of the former and 671.4 ft. from it the elevation is 41° 14'. Find the height of the balloon. Ans. 1000 ft. MISCELLANEOUS EXAMPLES 185 7. A ladder placed at an angle of 75° with the street just reaches the sill of a window 27 ft. above the ground on one side of the street. On turning the ladder over without moving its foot, it is found that when it rests against a wall on the other side of the street it is at an angle of 15° with the street. Find the breadth of the street. Ans. 3424 ft. 8. A man traveling due west along a straight road observes that when he is due south of a certain windmill the straight line drawn to a distant church tower makes an angle of 30° with the direction of the road. A mile farther on the bearings of the windmill and church tower are N.E. and N.W. respectively. Find the distances of the tower from the windmill and from the nearest point on the road. Ans. 2.39 mi., 1.37 mi. 9. Standing at a certain point, I observe the elevation of a house to be 45°, and the sill of one of its windows, known to be 20 ft. above the ground, sub- tends an angle of 20° at the same point. Find the height of the house. Ans. 54.94 ft. 10. A hill is inclined 36° to the horizon. An observer walks 100 yd. away from the foot of the hill, and then finds that the elevation of a point halfway up the hill is 18°. Find the height of the hill. Ans. 117.58 yd. 11. Two straight roads, inclined to one another at an angle of 60°, lead from a town A to two villages B and C ; B on one road distant 30 mi. from A, and C on the other road distant 15 mi. from A. Find the distance from B to C. Ans. 25.98 mi. 12. Two ships leave harbor together, one sailing N.E. at the rate of 7^ mi. an hour and the other sailing north at the rate of 10 mi. an hour. Prove that the distance between the ships after an hour and a half is 10.6 mi. 13. A and B are two positions on opposite sides of a mountain; C is a point visible from A and B. From A to C and from B to C are 10 mi. and 8 mi. respectively, and the angle BCA is 60°. Prove that the distance between A and Bis 9. 165 mi. 14. A and B are two consecutive milestones on a straight road and C is a distant spire. The angles ABC and BAG are observed to be 120° and 45° respectively. Show that the distance of the spire from A is 3.346 mi. 15. If the spire C in the last example stands on a hill, and its angle of ele- vation at A is 15°, show that it is .866 mi. higher than A. 16. If in Example 14 there is another spire D such that the angles DBA and DAB are 45° and 90° respectively and the angle DAC is 45°, prove that the distance from C to D is very nearly 2f mi. 17. A and B are consecutive milestones on a straight road ; C is the top of a distant mountain. At A the angle CAB is observed to be 38° 19'; at B the angle CBA is observed to be 132° 42', and the angle of elevation of C at B is 10° 15'. Show that the top of the mountain is 1243.7 yd. higher than B. 18. A base line AB, 1000 ft. long, is measured along the straight bank of a river; C is an object on the opposite bank; the angles BAC and CBA are observed to be 65° 37' and 53° 4' respectively. Prove that the perpendicular breadth of the river at C is 829.8 ft. 186 PLANE TRIGONOMETRY 19. The altitude of a certain rook is observed to be 47°, and after walking 1000 ft. towards the rook, up a slope inclined at an angle of 82° to the horizon, the observer finds that the altitude is 77°. Prove that the vertical height of the rock above the first point of observation is 1084 ft. 20. A privateer 10 mi. S.W. of a harbor sees a ship sail from it in a direc- tion S. 80° E., at a rate of 9 mi. an hour. In what direction and at what rate must the privateer sail in order to come up with the ship in 1^ hr. ? Ans. N. 76° 66' E. 13.9 mi. per hour. 21. At the top of a chimney 150 ft. high, standing at one corner of a tri- angular yard, the angle subtended by the adjacent sides of the yard are 80° and 46° respectively, while that subtended by the opposite side is 30°. Show that the lengths of the sides are 150 ft., 86.6 ft., and 106.8 ft. respectively. 22. A person goes 70 yd. up a slope of 1 in 3J- from the edge of a river, and observes the angle of depression of an object on the opposite bank to be 2J°. Find the breadth of the river. Ans. 422.18 yd. 23. A flagstaff h ft. high stands on the top of a tower. From a point in the plain on which the tower stands the angles of elevation of the top and bot- tom of the flagstaff are observed to be a and /3 respectively. Prove that the ,.,..,, . fttanfl .. . ftsinfl-cosar. height of the tower is - — ft., i.e. — : — - ft. tan a — tan /3 sin (a — /3) 24. The length of a lake subtends at a certain point an angle of 46° 24', and the distances from this point to the two extremities of the lake are 840 and 290 ft. Find the length of the lake. • Ans. 255.8 ft. 25. From the top of a cliff h ft. high the angles of depression of two ships at sea in a line with the foot of the cliff are a and /3 respectively. Show that the distance between the ships is h (cot/3 — cot or) ft. 26. Two ships are a mile apart. The angular distance of the first ship from a fort on shore, as observed from the second ship, is 35° 14' 10" ; the angular distance of the second ship from the fort, observed from the first ship, is 42° 11' 53" Find the distance in feet from each ship to the fort. Ans. 3121 ft., 3684 ft. 27. The angular elevation of a tower at a place due south of it is a, and at another place due west of the first and distant d from it, the elevation is /3. Prove that the height of the tower is d . dsina- sin/3 i.e. — Vcot 2 |3- cot 2 a Vsin (a — /3) • sin (a + /3) , 28. To find the distance of an inaccessible point C from either of two points A and B, having no instruments to measure angles. Prolong. CA to o, and OB to b, and join AB, Ab, and Ba. Measure AB, 500; aA, 100; aB, 560; bB, 100; and Ab, 550. Ans. 500 and 636. 29. A man stands on the top of the wall of height h and observes the angular elevation a of the top of a telegraph post ; he then descends from the wall and finds that the angular elevation is now /3 ; prove that the height of the post exceeds the height of the man by h — sin ((3 - a) MISCELLANEOUS EXAMPLES 187 30. Two inaccessible points A and B are visible from D, but no other point can be found whence both are visible. Take some point C, whence A and B can be seen, and measure CD, 200 ft.; ADC, 89°; ACD, 50° 30'. Then take some point E, whence D and B are visible, and measure DE, 200 ; BDE, 54° 30' ; BED, 88° 30'. At D measure ADB, 72° 30'. Compute the distance AB. Ans. 345.4 ft. 31. The angle of elevation of an inaccessible tower situated on a horizontal plane is 63° 26' ; at a point 500 ft. farther from the base of the tower the ele- vation of its top is 32° 14'. Find the height of the tower. Ans. 460.5 ft. 32. To compute the horizontal distance between two inaccessible points A and B, when no point can be found whence both can be seen. Take two points C and D, distant 200 yd., so that A can be seen from C, and B from D. From C measure CF, 200 yd. to F, whence A can be seen ; and from D measure DE, 200 yd. to E, whence B can be seen. Measure AFC, 83°; ACD, 53° 30' ; ACF, 54° 31'; BDE, 54° 30'; BDC, 156° 25'; DEB, 88° 30'. Ans. 345.3 yd. 33. A tower is situated on the bank of a river. From the opposite bank the angle of elevation of the tower is 60° 13', and from a point 40 ft. more distant the elevation is 50° 19'. Find the breadth of the Tiver. Ans. 88.9 ft. 34. A ship sailing north sees two lighthouses 8 mi. apart, in a line due west ; after an hour's sailing one lighthouse bears S.W. and the other S.S.W. Find the ship's rate. Ans. 13.6 mi. per hour. 35. A column in the north temperate zone is east-southeast of an observer, and at noon the extremity of its shadow is northeast of him. The shadow is 80 ft. in length, and the elevation of the column at the observer's station is 45°. Find the height of the column. Ans. 61.23 ft. 36. At a distance of 40 ft. from the foot of a tower on an inclined plane the tower subtends an angle of 41° 19' ; at a point 60 ft. farther away the angle sub- tended by the tower is 23° 45'. Find the height of the tower. Ans. 56.5 ft. 37. A tower makes an angle of 113° 12' with the inclined plane on which it stands ; and at a distance of 89 ft. from its base, measured down the plane, the angle subtended by the tower is 23° 27'. Find the height of the tower. Ans. 51.6 ft. 38. From the top of a hill the angles of depression of two objects situated in the horizontal plane of the base of the hill are 45° and 30° ; and the horizontal angle between the two objects is 30°. Show that the height of the hill is equal to the distance between the objects. 39. I observe the angular elevation of the summits of two spires which appear in a straight line to be a, and the angular depressions of their reflections in still water to be /3 and y. If the height of my eye above the level of the water be c, then the horizontal distance between the spires is 2 c cos'q sin (ft — 7) sin (j3 - or) sin (7 - a) 40. The angular elevation of a tower due south at a place A is 30°, and at a place B, due west of A and at a distance a from it, the elevation is 18°. Show a that the height of the tower is — - • V2V5+2 .188 PLANE TRIGONOMETRY 41. A boy standing c ft. behind and opposite the middle of a football goal sees that the angle of elevation of the nearer crossbar is A and the angle of elevation of the farther one is B. Show that the length of the field is c(tan.4 cot-B — 1). 42. A valley is crossed by a horizontal bridge whose length is I. The sides of the valley make angles A and B with the horizon. Show that the height of the bridge above the bottom of the valley is cot^l + cotB 43. A tower is situated on a horizontal plane at a distance a from the base of a hill whose inclination is a. A person on the hill, looking over the tower, can just see a pond, the distance of which from the tower is 6. Show that, if the distance of the observer from the foot of the hill be c, the height of the be sin a tower is a + 6 + c cos or 44. From a point on a hillside of constant inclination the angle of inclination of the top of an obelisk on its summit is observed to be a, and a ft. nearer to the top of the hill to be /3 ; show that if h be the height of the obelisk, the incli- nation of the hill to the horizon will be cos- x fa sin a: sin /3 "> \h sin(/3-a)J CHAPTEE X RECAPITULATION OF FORMULAS Plane Trigonometry Right triangles, pp. 2-1 1. (1) sin A = (2) cos A = (3) tan A = ^ (7) Side opposite an acute angle = hypotenuse x sine of the angle. (8) Side adjacent an acute angle = hypotenuse x cosine of the angle. (9) Side opposite an acute angle = adjacent side x tangent of the angle. Fundamental relations between the functions, p. 59. (19) sin x = (20) cos a; = (21) tana; = (22) tanx = cos X (23) sin 2 x + cos 2 x = 1. (24) sec 2 a; = 1 + tan 2 a;. csc a; 1 sec a; 1 cot X sin x csc x = sec x = cot x = SIM 1 cos a; 1 tana; cos x cot x = — sin x (25) csc 2 x = 1 + cot 2 x. Functions of the sum and of the difference of two angles, pp. 63-69. (40) sin (x + y) = sin x cos y + cos x sin y. (41) sin (x — y) = sin x cos y — cos x sin y. (42) cos (x + y) = cos x cos y — sin x sin y. (43) cos (a; — y) = cos a; cos y + sin x siny. 180 -y (44) tan (a; + y) = (45) tan (x — y) = (46) cot (x + y) = (47) cot (x — y) = 190 PLANE TRIGONOMETRY tan x + tan y 1 — tan x tan y tan x — tan y 1 + tan x tan y cot * cot ?/ — 1 cot y + cot x cot x cot y + 1 cot y — cot a; Functions of twice an angle, p. 70. (48) sin 2 x = 2 sin x cos a;. (49) cos 2 a; = cos 2 aj — sin 2 x. Functions of an angle in terms of functions of half the angle, p. 72. X X (51) sin x = 2 sin - cos - ■ Zl -J (52) cos a; = cos 2 -- ■ 2 •" (53) tan x 2tan- Zi 1 - tan 2 a; 2 Fund ions 01 . X sin- X cos ^ i half an angle, pp. 72-73- (58) (59) (54) — cos a; 2 tan- = (55) . Il + coscc - ± N 2 • cot ^ = 1 — COS : sin a; ■¥- ,, m . x ) l — cos a: x 1+cosa; (56) tan- = ±% Tw ^ (60) oot g =— j^-. (57) tan2 = -5»«_. (61) cot^^^. v y 2 1 +.cos x K ' 2 1 — cos x Sums and differences of functions, p. 74. (62) sin A + siaB = 2 sin $(A + B) cos £ (A — B). (63) sin A—sinB = 2 cos £ (4 + B) sin £ (^ - B). (64) cos4 +cos.B = 2cos£(J. + .B)cos£(,4 — B). (65) cos;! — cos£ = — 2 sin £(4 + £)sin J (4 — 5). .„„. sin^l +sin.B _ tan'^-(^ + B) ^ ' smA — sinB ~~ tan ^(4 — B) ' cos a; cos x RECAPITULATION OF FORMULAS 191 Law of sines, p. 102. a (72) sin A sinB sin C Law of cosines, p. 109. (73) a 2 = i 2 +c 2 - 2 be cos A. Law of tangents, p. 112. *• ' a — b tan^(A-B) Functions of the half angles of a triangle ia terms of the sides, pp. 113-115. s = I (a + b + e). (81) (82) (83) (84) sin \ A - -x ( * -m- 0) -\ be cos \ A - ->I4 ! — a) be tsm^A = ->l<* -b)(s- s(s — a) «) r=\l (s — a)(s — b) (s — c) s (85) tan£ A = s — a r (86) taniiJ = — - r (87) tan£C = s _ Area of a triangle, p. 117. (88) S = ^bc sin A. (89) S = Vs (« — a) (s — b) (s — c), SPHERICAL TRIGONOMETRY CHAPTEE I RIGHT SPHERICAL TRIANGLES 1. Correspondence between the face angles and the diedral angles of a triedral angle on the one hand, and the sides and angles of a spherical triangle on the other. Take any triedral angle O—A'B'C and let a sphere of any radius, as OA, be described about the vertex as a center. The intersections of this sphere with the faces of the triedral angle will be three arcs of great circles of the sphere, form- ing a spherical triangle, as ABC. The sides (arcs) AB, BC, CA of this triangle measure the face angles A' OB', B'OC", CO A' of the triedral angle. The angles ABC, BCA, CAB, are measured by the plane angles which also measure the diedral angles of the triedral angle; for, by Geometry, each is measured by the angle between two straight lines drawn, one in each face, perpendicular to the edge at the same point. Spherical Trigonometry treats of the trigonometric relations be- tween the six elements (three sides and three angles) of a spherical triangle ; or, what amounts to the same thing, between the face and diedral angles of the triedral angle which intercepts it, as shown in the figure. Hence we have the Theorem. From any property of triedral angles an analogous prop- erty of spherical triangles can be inferred, and vice versa. 193 194 SPHERICAL TRIGONOMETRY It is evident that the face and diedral angles of the triedral angle are not altered in magnitude by varying the radius of the sphere; hence the relations between the sides and angles of a spherical tri- angle are independent of the length of the radius. The sides of a spherical triangle, being arcs, are usually expressed in degrees.* The length of a side (arc) may be found in terms of any linear unit from the proportion circumference of great circle : length of arc : : 360° : degrees in arc. A side or an angle of a spherical triangle may have any value from 0° to 360°, but any spherical triangle can always be made to depend on a spherical triangle having each element less than 180°. Thus, a triangle such as ADEBC (unshaded portion of hemisphere in figure), which has a side ADEB greater than 180°, need not be considered, for its parts can be immediately found from the parts of the triangle ABC, each of whose sides is less than 180°. For arc ADEB = 360° — arc AB, angle CAD = 180° - angle CAB, etc. Only triangles whose elements are less than 180° are considered in this book. 2. Properties of spherical triangles. The proofs of the following properties of spherical triangles may be found in any treatise on Spherical Geometry: (a) Either side of a spherical triangle is less than the sum of the other two sides. (b) If two sides of a spherical triangle are unequal, the angles opposite them are unequal, and the greater angle lies opposite the greater side, and conversely. (c) The sum of the sides of a spherical triangle is less than 360°.f (d) The sum of the angles of a spher- ical triangle is greater than 180° and less than 540°. ± * One of the chief differences between Plane Trigonometry and Spherical Trigonometry is that in the former the sides of triangles are expressed in linear units, while in the latter alt the parts are usually expressed in units of arc, i.e. degrees, etc. t In a plane triangle the sum of the sides may have any magnitude. t In a plane triangle the sum of the angles is always equal to 180°, EIGHT SPHERICAL TRIANGLES 195 (e) If A'B'C is the polar triangle* of ABC, then, conversely, ABC is the polar triangle of A'B'C (/) In two polar triangles each angle of one is the supplement of the side lying opposite to it in the other. Applying this to the last figure, we get A = 180° - a', B = 180° - V, C = 180° - c', A' = 180° -a, B' = 180°-b, C = 180° - c. A spherical triangle which has one or more right angles is called a right spherical triangle. EXAMPLES 1. Find the sides of the polar triangles of the spherical triangles whose angles are as follows. Draw the figure in each case. (a) A = 70°, B = 80°, C = 100°. Arts, a' = 110°, V = 100°, c' = 80°. (b) A = 56°, B = 97°, C = 112°. (c) A = 68° 14', B = 52° 10', C = 98° 44'. (d) A = 115.6°, B = 89.9°, C = 74.2°. 2. Find the angles of the polar triangles of the spherical triangles whose sides are as follows: (a) a = 94°, 6 = 52°, c = 100°. Ans. A' = 86°, B' = 128°, C = 80°. (b) a = 74° 42', 6 = 95° 6', c = 66° 25'. (c) a = 106.4°, 6 = 64.3°, c = 51.7°. 3. If a triangle has three right angles, show that the sides of the triangle are quadrants. 4. Show that if a triangle has two right angles, the sides opposite these angles are quadrants, and the third angle is measured by the opposite side. 5. Find the lengths of the sides of the triangles in Example 2 if the radius of the sphere is 4 ft. 3. Formulas relating to right spherical triangles. Erom the above Examples 3 and 4, it is evident that the only kind of right spherical triangle that requires further investigation is that which contains only one right angle. In the figure shown on the next page let ABC be a right spherical triangle having only one right angle, the center of the sphere being at O. Let C be the right angle, and suppose first that each of the other efemenAl is less than 90°, the radius of the sphere being unity. . le |rt *Th&poZ«r Irwhgle of any spherical triangle is constructed by describing arcs of great circles aboi^tfthe jjrtices of the original triangle as poles. 196 SPHERICAL TRIGONOMETRY Pass an auxiliary plane through. B perpendicular to OA, cutting 0.4 at E and OC at £>. Draw BE, BD, and BE. BE and DE are each perpendicular to OA ; [If a straight line is J_ to a plane, it is _L to every line in trie plane.] N therefore angle BED = angle A. The plane BDE is perpendicular to the plane AOC ' [" If a straight line is J_ to a plane, every plane"! Lpassed through the line is X to the first plane. J hence BD, which is the intersection of the planes BDE and BOC, is perpendicular to the plane AOC, ["If two intersecting planes are each X to a third"] . Lplane, their intersection is also X to that plane.J and therefore perpendicular to OC and DE. (&V t In triangle EOD, remembering that angle EOD = b, we have 07? OD = cos b, ,Si <: ! or, clearing of fractions, {A) OE = OD ■ cos b. But OE = cos o (= cos EOB), and OD = cos a (== cos DOB). Substituting in (,1), we get (1) cos c — cos a cos b. In triangle BED, remembering that angle BE D = angle A, we have BD . = SHI yl , BE ' or, clearing of fractions, (23) BD =BE- sin A. But BD = sin a (= sin DOB), and BE = sin c (= sin EOB). RIGHT SPHERICAL TRIANGLES 197 Substituting in (B), we get (2) sin a = sin c sin A. Similarly, if we had passed the auxiliary plane through A perpen- dicular to OB, (3) sin b = sin c sin B. Again, in the triangle BED, (O cos ^=ff- DE But DE = OD sin b, from sin b = OD OD = cos a (= cos DOB), and .B.E = sin c (= sin £0.8). Substituting in (C), ,_,. ODsinS sinJ (D) cos .4 = — ; = cos a ■ —. sin c sin c But from (3), — — = sin B. Therefore v '' sine (4) cos A — cos a sin B. Similarly, if we had passed the auxiliary plane through A perpen- dicular to OB, (5) cos B — cos 6 sin A. The above five formulas are fundamental ; that is, from them we may derive all other relations expressing any one part of a right spherical triangle in terms of two others. For example, to find a relation between A, b, c, proceed thus : Erom (4), cos A = cos a sin B _ cos c sin b cos b sin e [Since cos a = ^^ from (1), and sin B= — — from (3). cos b sm c J sin b cos c cos b sin c (6) .*. cos A = tan b cot c. Similarly, we may get (7) COSB: = tan a cot c. (8) sin b ■. = tan a cot A. (9) sin a : = tan 6 cot B. (10) COSC : = cot A cot B. 198 SPHERICAL TRIGONOMETRY These ten formulas are sufficient for the solution of right spher- ical triangles. In deriving these formulas we assumed all the elements except the right angle to be less than 90°. But the formu- las hold when this assumption is not made. ' For instance, let us suppose that a is greater that 90°. In this case the auxiliary plane BDE will cut CO and AO produced beyond the center 0, and we have, in triangle EOD, OE (E) cos DOE (= cos b) = But and OD OE = cos EOB = — cos A OB = — cos c, OD = cos DOB = — cos COB — — cos a. Substituting in (E), we get cos o , cos b = > or cos o = cos a cos o, cos a which is the same 8T (1). Likewise, the other formulas will hold true in this case. Similarly, they may be shown to hold true in all cases. If the two sides including the right angle are either both less or o 6 both greater than 90° (that is, cos a and cos b are either both positive or both negative), then the product (F) cos a cos b will always be positive, and therefore cose, from (1), will always be positive, that is, c will always be less than 90°. If, however, one of the sides including the right angle is less and the other is greater than 90°, the product (F), and therefore also cos c, will be negative, and c will be greater than 90°. Hence we have Theorem I. If the two sides including the right angle of a right spherical triangle are both less or both greater than 90°, the hypotenuse EIGHT SPHEEIGAL TRIANGLES 199 is less than 90° ; if one side is less and the other is greater than 90°, the hypotenuse is greater than 90°. -T7. ,,, t ,„, • „ cos A ., . , cos-B From (4) and (5), sinB = > and smi = r- - x ' ' cos a eoso Since A and B are less than 180°, sin A and sinB mnst always be positive. But then cos A and cos a must have the same sign, that is, A and a are either both less than 90° or both greater than 90°. Simi- larly, for B and b. Hence we have Theorem II. In a right spherical triangle an oblique angle and the side opposite are either both less or both greater than 90°. 4. Napier's rules of circular parts. The ten formulas derived in the last section express the relations between the three sides and the two oblique angles of a right spherical triangle. All these relations may be shown to follow from two very useful rules discovered by Baron Napier, the inventor of logarithms. For this purpose the right angle (not entering the formulas) is not taken into account, and we replace the hypotenuse and the two B B, b b oblique angles by their respective complements ; so that the five parts, called the circular parts, used in Napier's rules are a, b, A c , c c , B c . The subscript c indicates that the complement is to be used. The first figure illustrates the ordinary method of represent- ing a right spherical triangle. To emphasize the circular parts employed in Napier's rules, the same triangle might be represented as shown in the second figure. It is not necessary, however, to draw the triangle at all when using Napier's rules ; in fact, it is found to be more convenient to simply write down the five parts in their proper order as on j^ s e the circumference of a circle, as shown in the third figure (hence the name circular parts). , Any one of these parts may be called a middle part; then the two parts immediately adjacent to it are called adja- cent parts, and the other two opposite parts. Thus, if a is taken as a middle part, B c and b are the adjacent parts, while c c and A c are the opposite parts. 200 SPHERICAL TRIGONOMETRY Napier's rules of circular parts. Rule I. The sine of any middle part is equal to the product of the tangents of the adjacent parts. Rule II. The sine of any middle part is equal to the product of the cosines of the opposite parts. These rules are easily remembered if we associate the first one with the expression " tan-adj." and the second one with " cos-opp." * Napier's rules may be easily verified by applying them in turn to each one of the five circular parts taken as a middle part, and com- paring the results with (1) to (10). For example, let c c be taken as a middle part ; then A c and B c are the adjacent parts, while a and b are the opposite parts. Then, by Rule I, gin ^ = Un ^ tan ^ c ° or, cos c = cot A cot B ; A c b c which agrees with (10), p. 197. By Rule II, sin c,. = cos a cos b, I, a or, cos c = cos a cos b ; which agrees with (1), p. 196. The student should verify Napier's rules in this manner by taking each one of the other four circular parts as the middle part. Writers on Trigonometry differ as to the practical value of Napier's rules, but it is generally conceded that they are a great aid to the memory in applying formulas (1) to (10) to the solution of right spherical triangles, and we shall so employ them. 5. Solution of right spherical triangles. To solve a right spherical triangle, two elements (parts) must be given in addition to the right angle. For the sake of uniformity we shall continue to denote the right angle in a spherical triangle ABC by the letter C. General directions for solving right spherical triangles. £ b -y- First step. Write down the five circular parts as in first figure. Second step. Underline the two given parts and the required un- known part. Thus, if A c and a are given to find b, ive underline all three as is shown in the second figure. * Or by noting that a is the first vowel in the words " tangent'* and " adjacent," while o is the first vowel in the words " cosine " and " opposite." EIGHT SPHERICAL TRIANGLES 201 Third step. Pick out the middle part (in this case b) and cross the line under it as indicated in the third figure. Fourth step. Use Rule I if the other tivo parts are adjacent to the middle part (as m case illustrated), or Rule II if they are opposite, and solve for the unknoivn part. Check : Check with that rule which involves the three required parts* Careful attention must be paid to the algebraic signs of the func- tions when solving spherical triangles ; the cosines, tangents, and cotangents of angles or arcs greater than 90° being negative. When computing with logarithms we shall write (n) after the logarithms when the functions are negative. If the number of negative factors is even, the result will be positive ; if it is odd, the result will be negative and (n) should be written after the resulting logarithm. In order to be able to show our computations in compact form, we shall write down all the logarithms of the trigonometric functions just as they are given in our table ; that is, when a logarithm has a negative characteristic we will not write down — 10 after it. f Ex. 1. Solve the right spherical triangle, having given B - Solution. Follow the above general directions. 33° 50', a = 108°. To find A c c A& Using Rule II sin .4.,, = cos B c cos a cos A = sin B cos a log sin 5 = 9.7457 log cos a = 9.4900 (n) log cosA = 9.2357 (n) .-. 180°-A±=80°6' and A = 99° 54'. To find b n, : b_ a Using Rule I sin a = tan B c tan 6 tan b = sin a tan B log sin a = 9.9782 log tan B= 9.8263 log tan 6 = 9.8045 .-. 6 = 32° 31'. To find c Cc $ b Using Rule I sin B c = tan c c tan a cote = COS.B cot a log cos# = 9.9194 log cot a = 9.5118 (n) log cot c = 9.4312 (n) . . 180° - c = 74° 54' and c = 105° 6'. The value of log cos A found is the same as that found in our first computa- tion. The student should observe that in checking our work in this example * Thus, in above case, Ac and a are given ; therefore we underline the three required parts and cross b as the middle part. Applying Rule II, cc and B e being opposite parts, we get sin b = cos Cc cos B e , or, sin 6 = sin cein.fi. t For example, as in the table, we will write log sin 24° = 9.6093. To be exact, this should be written log sin 21° = 9.6093 -10, Or, logsin24°= f.6093. + Since cos A is negative, we get the supplement of A from the table. 202 SPHERICAL TRIGONOMETRY c c it was not necessary to look up any new logarithms. Hence the check in this case is only on the correctness of the logarithmic work.* Check: Using Rule I sin .A,, = tan 5 tan c„ A . jj cos A = tan b cote ~^ C • log tan 6 = 9.8045 log cot c = 9.4312 (n) b_ a log cos^l = 9.2357 (n) In logarithmic computations the student should always write down an outline or skeleton of the computation before using his logarithmic table at all. In the last example this outline would be as follows : log sin B = log cos o = log cos A = .: 180° -A = and A = (n) (n) log sin a = log tan B = log tan 6 = .-. 6 = log cos B = log cot a = log cot c = .. 180°- c = and c = (71) (n) It saves time to look up all the logarithms at once, and besides it reduces the liability of error to thus separate the theoretical part of the work from that which is purely mechanical. Students should be drilled in writing down forms like that given above before attempt- ing to solve examples. Ex. 2. Solve the right spherical triangle, having given c = 70° 30', A = 100°. Solution. Follow the general directions. To find a To find b To find B b a -r- Using Rule II sin a = cos c c cosmic sin a = sin c sin A log sine = 9.9743 log sin A = 9.9934 log sin a = 9.9677 .-. 180°- at =68° 10' and a = 111° 50'. * J3o b_ Using a Rule I sin.A = tanb = log cos A - log tan c = : tan 6 tan c c = cos A tan c = 9.2397 (») = 0.4509 log tan 6 = 9. 6906 (n) .-. 180° -6 = 26° 8' and 6 = 153° 52'. 6 o Using Rule I sin c c = tan A c tan B c cot B = cos c tan A log cose = 9.5235 log tan^L = 0.7537 (n) log cot B = 0.2772 (n) .-. 180°-B=27°51' and B = 152° 9'. The work of verifying the results is left to the student. * In order to be sure that the angles and sides have been correctly taken from the tables, in such an example as this, we should use them together with some of the given data in relations not already employed. t Since a is determined from its sine, it is evident that it may have the value 68° 10' found from the table, or the supplementary value 111 50' . Since A > 90°, however, we know from Th. II, p. 199, that a > 90° ; hence a= 111 50' is the only solution. EIGHT SPHERICAL TRIANGLES 203 6. The ambiguous case. Two solutions. When the given parts of a right spherical triangle are an oblique angle and its opposite side, there are two triangles which satisfy the given conditions. For, in the triangle ABC, let C = 90°, and let .4 and CB ,i<^ J J' (= a) be the given parts. If we extend AB and AC to A', it is evident that the triangle A'BC also satisfies the given condi- tions, since BCA' = 90°, A' = A, and BC = a. The remaining parts in A'BC are supplementary to the respective remaining parts in ABC. Thus A'B = 180°- c, A'C = 180°- b, A'BC = 180° -ABC. This ambiguity also appears in the solution of the triangle, as is illustrated in the following example : Ex. 3. Solve the bright spherical triangle, having given A = 105° 59', a = 128° 33'. Solution. We proceed as in the previous examples. To find b sin 6 = sin 6 = a_ tan a tan^. c tan a cot A log tan a = 0.0986 (n) iogcot^L= 9.4570 (n) log sin 6 = 9.5556 .-. b = 21° 4', or, 180°- 6 =158° 56' = 6'.* To find B Cc 4-° sin A c = cos a cos B c . „ cos A sin B = cos a log cos A = 9.4399 (n) log cos a = 9.7946 (n) log sin B = 9.6453 .-. B = 26° 14', or, 180° -B =153° 46'= B'. To find c A c b a sin a = cos^l,, cos c„ sin a sm c = ■ smA log sin a = 9.8932 log sin^l = 9.9828 log sine = 9. 9104 .-. c'=54°27', or, 180°-c'=125°33'=c.t Hence the two solutions are : 1. 6 = 21° 4', c = 125° 33', B = 26° 14' (triangle ABC) ; 2. V = 158° 56', . c' = 54° 27', B' = 153° 46' (triangle A'BC). It is not necessary to check both solutions. We leave this to the student. * Since sin B is positive and B is not known, we cannot remove the ambiguity. Hence both the acute angle taken from the table and its supplement must be retained. t The two values of B must be retained, since b has two values which are supplementary- t Since a > 90° and b has two values, one > and the other < 90°, it follows from Th. I, p. 198, that c will have two values, the first one < 90° and the second > 90°. 204 SPHEKICAL TEIGONOMETEY EXAMPLES Solve the following right spherical triangles : No. Given Parts Keqdieed Parts 1 a = 132° 6' 6 = 77° 51' A = 131° 27' B = 80° 55' c = 98° 7' 2 a = 159° c = 137° 20' A = 148° 5' B = 65° 23' 6 = 37° 54' 3 A = 50° 20' B = 122° 40' a = 40° 42' 6 = 134° 31' c = 122° V 4 a = 160° 6 = 38° 30' A = 149° 41' B = 66° 44' c = 137° 20' 5 £ = 80° 6 = 67° 40' 4 = 27° 12' a = 25° 25' c = 69°54'; or, A'= 152° 48' a'= 154° 35' c'= 110° 6' 6 B = 112° c = 81° 50' A = 109° 23' a = 110° 58' 6 = 113° 22' 7 a = 61° B = 123° 40' A = 66° 12' 6 = 127° 17' c = 107° 5' 8 a = 61° 40' b = 144° 10' 4 = 72° 29' B = 140° 38' c = 112° 38' 9 A = 99° 50' a = 112° 5 = 27° 7' b = 25° 24' c= 109° 46'; or, B'= 152° 53' &'= 154° 36' c'= 70° 14' 10 6 = 15° c = 152° 20' A = 120° 44' a = 156° 30' B = 33° 53' 11 A = 62° 59' B = 37° 4' a = 41° 6' 6 = 26° 25' c = 47° 32' 12 A = 73° 7' c = 114° 32' a = 60° 31' B = 143° 50' b = 147° 32' 13 B = 144° 54' 6 = 146° 32' 4 = 78° 47' a = 70° 10' c = 106°28'; or, A'= 101°. 13' a'= 109° 50' c'= 73° 32' 14 B = 68° 18' c = 47° 34' A = 30° 32' a = 22° 1' 6 = 43° 18' 15 A = 161° 52' b = 131° 8' a = 166° 9' .B = 101° 49' c = 50° 18' 16 a = 113° 25' 6 = 110° 47' X = 112° 3' B = 109° 12' c = 81° 54' 17 a = 137° 9' B = 74° 51' A = 135° 3' 6 = 68° 17' c = 105° 44' 18 .4 = 144° 54' B = 101° 14' a = 146° 33' 6 = 109° 48' c = 73° 35' 19 a = 69° 18' c = 84° 27' ^ = 70° B = 75° 6' 6 =74° 7' 20. For more examples take any two parts in the above triangles and solve for the other three. 7. Solution of isosceles and quadrantal triangles. Plane isosceles triangles -were solved by dividing each one into two equal right tri- angles and then solving one of the right triangles. Similarly, we may solve an isosceles spherical triangle by dividing it into two sym- metrical right spherical triangles by an arc drawn from the vertex perpendicular to the base, and then solving one of the right spheri- cal triangles. A quadrantal triangle is a spherical triangle one side of which is a quadrant (= 90°). By (/), p. 195, the polar triangle of a quad- rantal triangle is a right triangle. Therefore, to solve a quadrantal triangle we have only to solve its polar triangle and take the sup- plements of the parts obtained by the calculation. Ex. 1. Solve the triangle, having given c = 90°, a = 07° 38', b = 48° 50'. Solution. This is a quadrantal triangle since one side c = 90°. We then find the corresponding elements of its polar triangle by (/), p. 195. They are C'=90°, A' = 112° 22', B' = 131° 10'. We solve this right triangle in the usual way. EIGHT SPHERICAL TRIANGLES 205 Construct the polar (right) triangle. Given A' = 112° 22', B' = 131° 10': To find a' Cc' 4 # sin 4 ' c = cos a' cos .Be . . cos A' cos a = sinB' log cos 4'= 9.5804- (n) log sin B' =08767 log cos a' =9.7037 (n) 180° - a' = 59° 38'. a' = 120° 22'. Similarly, we get 6' = 135° 23', c' = 68° 55'. Hence in the given quadrantal tri- angle we have A = 180° - a' = 5C° 38', B = 180° - V = 44° 37', G = 180° - c' = 111° 5'. EXAMPLES Solve the following quadrantal triangles : No. Given Parts Required Parts 1 A = 139° 6 = 143° c = 90° a = 117° 1' JS = 153° 42' C = 132° 34' 2 A = ib° 30' B = 139°20' c = 90° a = 57° 22' 6 = 129° 42' C = 57°53' 3 a = 30° 20' C = 42°40' c = 90° A = 20°V £ = 141° 30' 6 = 113° 17' 4 B=70°12' C=106°25' c = 90° A = 33° 28' a = 35° 4' 6 = 78° 47' 5 A = 105° 53' a = 104° 54' c = 90° B = 69°16' 6 = 70° C = 84°30'; or B=110°44' 6 = 110° C = 95°30' Solve the following isosceles spherical triangles : No. Given Parts Required Parts 6 a = 54° 20' c = 72° 54' A = B 6 = 54° 20' A = B = 57° 59' C = 93° 59' 7 a = 54°30' C = 71° A = B b = 54° 30' A = B = 67° 30' c = 56° 26' 8 a = 66°29' A = B = 50°1T 6 = 66° 29' c=lll°30' C = 128°42' 9 c = 156°40' C = 187°46' A = B a = b = 79° or 101° A = B = 199° 34' Prove the following relations between the elements of a right spherical triangle (O = 90°): 10. eosMsin 2 e = sin(e + a)sin(c — a). 13. sin (6 + c) = 2 cos 2 ^ A cos 6 sin c. 11. tan a cos c = sin 6 cot B. 14. sin (c — 6) = 2 sin 2 £ 4 cos 6 sin c. 12. sin 2 X = cos 2 B + sin 2 a sin 2 B. CHAPTEE II OBLIQUE SPHERICAL TRIANGLES 8. Fundamental formulas. In this chapter some relations between the sides and angles of any spherical triangle (whether right angled or oblique) will be derived. 9. Law of sines. In a spherical triangle the sines of the sides are proportional to the sines of the opposite angles. Proof. Let ABC be any spherical triangle, and draw the arc CD perpendicular to AB. There will be two cases according as CD falls C upon AB (first figure) or upon AB produced (second figure). For the sake of brevity let CD =p, AD = n, BD = m, angle ACD = x, angle BCD — y. In the right triangle ADC (either figure) (A) sis. p = sin b sin A. In the right triangle BCD (first figure) (B) sinp == sin a sin B. This also holds true in the second figure, for sin DBC = sin (180° -B) = sinB, Equating the values of sinp from (A) and (B), sin a sinB = sin b sin A, or, dividing through by sin^i sinB, sin a sin b Eule II, p. 200 Eule II, p. 200 (C) sin A sinB 206 OBLIQUE SPHERICAL TRIANGLES 207 In like manner, by drawing perpendiculars from A and B, we get , „ sin b sine (D) - T — = - T —- , and x / Gin R cin fl sin B sin C sin c _ sin a sinC sin.4' . . sine sin a ,. , ( E) —. — - = — — 7 , respectively. Writing (C), (D), (E) as a single statement, we get the law of sines, . sin a sin b sin c * ^ sin.4 — sinl? — sinC 10. Law of cosines. In a spherical triangle the cosine of any side is equal to the product of the cosines of the other two sides plus the product of the sines of these two sides and the cosine of their included angle. Proof. Using the same figures as in the last section, we have in the right triangle BDC, cos a = cosp cos m Rule II, p. 200 = cos p cos (c — n) = cosp { cos c cos n + sin c sin n j (A) = cosp cos c cos n + cosp sin c sin n. In the right triangle ADC, (B) cosp cos n = cos b. ' cos b Whence cos p = > cosn and, multiplying both sides by sin n, cos b , . (C) cospsmn = sin n = cos b tan n. K ' ± cos n But cos A = tan n cot b, or, Rule I, p. 200 ( D) tan n = tan b cos A. Substituting value of tan re from (D) in (C), we have . (E) cosp sin n = cos b tan b cos A = sin boos A. Substituting the value Of cosp cos re from (B) and the value of cosp sin re from (E) in (A), we get the law of cosines, (F) cos a = cos b cos c + sin 5 sin c cos ^1. » Compare with the law of sines in Granville's Flame Trigonometry, p. 102. 208 SPHERICAL TRIGONOMETRY Similarly, for the sides b and c we may obtain (£) cos b = cos c cos a + sin c sin a cos B, (fl) cos c = cos a cos b + sin a sin b cos C. 11. Principle of Duality. Given any relation involving one or more of the sides a, b, c, and the angles A, B, C of any general spherical triangle. Now the polar triangle (whose sides are denoted by a', b', c', and angles by A', B', C") is also in this case a general spherical triangle, and the given relation must hold true for it also ; that is, the given relation applies to the polar triangle if accents are placed upon the letters representing the sides and angles. Thus (F), (G), (Zf) of the last section give ns the following law of cosines for the polar triangle : (A) cos a' = cos b' cos c' + sin V sin c' cos A '. (E) cos V = cos c' cos a' + sin c' sin a' cos B'. (C) cos c' = cos a' cos b' + sin a' sin b' cos C. But by (/), p. 195, a' = 180°-^, b' = 180°-B, c' = 180°-C, A' = 180°-a, B' = 180°-b, C" = 180°-c. Making these substitutions in (^4), (B), (C), which refer to the polar triangle, we get (D) cos (180° - A) = cos (180° - B) cos (180° - C) + sin (180° - B) sin (180° - C) cos (180° - a), (E) cos (180° - B) = cos (180° - C) cos (180° - A ) + sin (180° - C) sin (180° - A) cos (180° - b), (F) cos (180° - C) = cos (180° - A) cos (180° - B) + sin (180° - A) sin (180° - B) cos (180° - c), which involve the sides and angles of the original triangle. The result of the preceding discussion may then be stated in the following form : Theorem. In any relation between the parts of a general spherical triangle, each pari may be replaced by the supplement of the opposite part, and the relation thus obtained will hold true. OBLIQUE SPHERICAL TKIANGLES 209 The Principle of Duality follows when the above theorem is applied to a relation involving one or more of the sides and the supplements of the angles (instead of the angles themselves). Let the supplements of the angles of the triangle be denoted by a, /?, y * ; that is, a = 180°- A, j8 = 180°-£, y = 180°-C, or, A =180°- a, B = 180°-p, C = 180°-y. When we apply the above theorem to a rela- tion between the sides and supplements of the angles of a triangle, we, in fact, replace a by a (= 180° — A), replace b by /3 (= 180° -B), replace e by y (= 180° - C), replace a (=180° -A) by 180° -(180°- a)=a, replace £ (= 180° -B) by 180°- (180°- b) = b, replace y (=180°-C) by 180° -(180°- c) = c, or, what amounts to the same thing, interchange the Greek and Soman letters. For instance, substitute A =180° -a, B = 180°- ft C = 180°-y in (F), (G), (H) of the last section. This gives the law of cosines for the sides in the new form (12) (13) (14) cos a = cos b cos c — sin b sin c cos a, cos b = cos c cos a — sin c sin a cos jG, cos c = cos a cos b — sin a sin b cos y. [Since cos A = cos (180° — <*)*= - cos a, etc.] If we now apply the above theorem to these formulas, we get the law of cosines for the angles, namely, (15) cos a = cos /? cos y — sin /? sin y cos a, (16) cos fl = cos y c° 8 a — sin y sin a cos b, (17) cos y = cos a cos /? — sin a sin cos c, * a, /3, y are then the exterior angles of the triangle, as shown in the figure. 210 SPHERICAL TRIGONOMETRY that is, we have derived three new relations between the sides and supplements of the angles of the triangle.* We may now state the Principle of Duality. If the sides of a general spherical triangle are denoted by the Roman letters a, b, c, and the supplements of the cor- responding opposite angles by the Greek letters a, /?, y, then, from any given formula involving any of these six parts, we may write down a dual formula by simply interchanging the corresponding Greek and Roman letters. The immediate consequence of this principle is that formulas in Spherical Trigonometry occur in pairs, either one of a pair being the dual of the other. Thus (12) and (15) are dual formulas ; also (13) and (16), or (14) and (17). If we substitute A = 180°- a, B = 180°- ft C = 180°-y in the law of sines (p. 207), we get sin a _ sin b sin c sin a sin /? sin y [Since sin A = sin (180° — a) = sin a, etc.] Applying the Principle of Duality to this relation, we get sin a _ sin /3 _ sin y sin a sin b sin c which is essentially the same as the previous form. The forms of the law of cosines that we have derived involve algebraic sums. As these are not convenient for logarithmic calcu- lations, we will reduce them to the form of products. 12. Trigonometric functions of half the supplements of the angles of a spherical triangle in terms of its sides. Denote half the sum of the sides of a triangle (i.e. half the perimeter) by s. Then (A) 2s = a + b + c, or, s = ^(a + b + c). * If we had employed the interior angles of the triangle in our formulas (as has been the almost universal practice of writers on Spherical Trigonometry), the two sets of cosine formulas would not have been of the same form. That the method used here has many advantages will become more and more apparent as the reading of the text is continued. Not only are the resulting formulas much easier to memorize, but much labor is saved in that, when'we have derived one set of formulas for the angles (or sides), the corresponding set of formulas for the sides (or angles) may be written down at once by mere inspection by applying this Principle of Duality. The great advantage of using this Principle of Duality was first pointed out by MSbius (1790-1868). OBLIQUE SPHERICAL TRIANGLES 211 Subtracting 2 c from both sides of (^4), 2s — 2c = a + b + c — 2 c, or, (B) s-c = ) s (a + b-c). Similarly, (C) s-b = $(a-b + c), and (-D) « -« = $(- a + 5 + c) = £(Z» + c - a). From Plane Trigonometry, (E) 2 sin 2 J a = 1 — cos a, (F) 2 cos 2 £ a = 1 + cos a. But from (12), p. 209, solving for cos a, cos b cos c — cos a cos a = hence (E) becomes 2sin 2 i sin b sin c 1- cos 5 cos c — cos a sin b sin c sin J sin c — cos 5 cos c + cos a sin & sin c cos a — (cos b cos c — sin b sin c) sin 5 sin c _ cos a — cos (b + c) sin 6 sin o _ — 2 sin £ (a + 6 + e) sin £ (a — 5 — c) # or, sin 6 sin c (G) 2sinH« = 2sin ^ + & + c)sill2 - (& + C - a) - v sin 5 sin c [Since sin § (a — b — c)= — sin \(—a + b + c)= — sin £(& + c — a).] Substituting from (4) and (Z>) in ((?), we get i 2 i sin s sin (s — a) sin" A- a = : — r^ » or, J sin 6 sin c ,,„. . , I sins sin (s — a) (18) sin|a = >J ^-r-\ '- K ' \ sin b sine *Let -4 = a .4=a £=6+c B=b+c A+B=a+b+c A—B=a—b—c l(A + B)=t(a + b + c). i(A-B)=i(a-b-c). Hence, substituting in (65), P. 74, Granville's Plane Trigonometry, namely, cos .4 — cos B= — 2Bini(A + B)sin^{A-B), we get cos a — cos (6 + c)= — 2 sin i (a + b + c) sin J(a - b - c). 212 SPHERICAL TRIGONOMETRY Similarly, (F) becomes . . , cos b cos o — cos a 2 cos 2 A « = H ^-r— : 2 sin b sin c , sin 6 sin c + co s & cos c — cos a sin b sin c _ cos, (b — o) — cos a sin 6 sin c — 2 sin 1 (a + b — o) sin A (& — c — a) * = — : ^ *-^ > or, sin 6 sin c 2 sin i- (a + b — c) sin A (a — b + c) (H) 2cos 2 Aa = * K . / . — ^ z - v ' J sin b sin o [Since sin J (b - c - a) = - sin J (- b + c + a) = - sin J (a - b + c).] Substituting from (B) and (C) in (H), we get sin (s — c) sin (s — b) cos 2 A« = * — : — '— -. — 1 or, sin b sm c , , , /sin (s — &) sin (s — c) (19) cosia=-<J i — . I .. -• v ' \ sin 6 sin c sin -i- # Since tan \a = -—; we get from this, by substitution from (18) and (19), (20) T sin s sin (s — a) f tania=-Aj-r-7 , N \ , } ■' \ sin (s — b ) sin ( s — c) * Let ^4=&-c A=b-c B=a B= a A+B=a+b— c A — B= b—c—a ^A + B)=^(a+b-c). i(^-B)=i(6-c-a). Hence, substituting in formula (65), found on p. 74, Granville's Plane Trigonometry, namely, cos A - cos B= - 2 sin £(^1 + B) sin J(^4 - .B), we get cos(fc-c) — cosa=-2sin£(ct + &-c)sin£(6-c — a). t In memorizing these formulas it will be found an aid to the memory to note the fact that under each radical (a) only the sine function occurs. (b) The denominators of the sine and cosine formulas involve those two sides of the tri- angle which are not opposite to the angle sought. (c) "When reading the numerator and denominator of the fraction in the tangent formula, £ comes first and then the differences s — a, s-6, s — c, in cyclical order ; s and the first difference occurring also in the numerator of the cor- responding sine formula, while the last two differences occur in the numerator of the corresponding cosine formula. OBLIQUE SPHERICAL TRIANGLES 213 In like manner, we may get (21) sinl^ = - N j: sin s sin (s — V) sin c sin a (22) cosf/?=^j!i sin (s — c) sin (s — a) sin c sin a <*> -^H jr ,in(s " 6) : Also » sin (s — c) sin (s — a _) .„.. . , I sin s sin (s — c) ( 24 ) sin ly = 'V : . ,. ■ v ' * r N sin a sin b (25) cos|y = j!^iZLf)i^li), ' > sin a sin 6 .„„. , sin s sin (s — c) v y 2r \sin(s — a)sin(s — 6) In solving triangles it is sometimes more convenient to use other forms of (20), (23), (26). Thus, in the right-hand member of (20), multiply both the numerator and denominator of the fraction under the radical by sin(s — a). This gives r sm s sm 2 (s — a) tania; = A 2 Ns i sin (s — a) sin (*■ — V) sin (s — o) . \_~^ sins = sin (s—a) '<J -^— v ' >sin(s— o a) sin (s— b)ein (s—c) # _ Isin (s — a) sin (s — b) sin (s — c) Let tan A d 1 y sm s sin (s — a) then tan \a= ,_ , , • 2 tan \ a Similarly, for tan \ /? and tan \ y. Hence , . |sin(s — a)sin(s— 6) sin (s — c) (27) tan|^>J-i l s l (28) tan | a = sin (s — a) tan|d » -w-tS^ (30) tan|y = 2 ' sin (s — c) tan|d ■• It may be shown that d = diameter of the circle inscribed in the spherical triangle. 214 SPHEKICAL TRIGONOMETRY 13. Trigonometric functions of the half sides of a spherical triangle in terms of the supplements of the angles. By making use of the Princi- ple of Duality on p. 208, we get at once from formulas (18.) to (30), by replacing the supplement of an angle by the opposite side and each side by the supplement of the opposite angle, the following formulas : (33) (42) (43) , , . , Isina sin (a — a) (31) sin§a = -0 —± L, v ' \ sin/? sin y (32) cos!a = -y sin (<y — p) sin (<r — y) sin /? sin y sin o" sin (a — a) tan I a = \H— -, „ s . ; 2 Nsin(o-— B)sinCo- — (er-/?)sin(o--y)' ,„.-. . , . Isin a sin (a — B) (34) sin | b = aJ : ^ ^ , v ' > sin y sin a (35) cos | b = Jsin(o--y)sin(o--a) s ' > sin v sin a /„„n i , I sin a sin (a — B) <*> tan ' &= N sin(o--y)sin(a-a) ' (37) sin| C=A J Sin<rsin ^E l), K ' y sin a sin p (38) cos|c=^ sin (a — a) sin (<r — /?) sin a sin fj (39\ ■ tan 1 c - i l ~~ sin °" Sin ( °" ~ y ~ (39) tan , c _ \ sin((r _ a)sin((r _ ^ • (40) tanlg* => j!^- a)sin ( a -^ S in ^-y) /.-v i sin(o-— a) (41) tan|a= v J , K ' tan Jo smcr sin (a- — /?) tan|6= /■ /' . tan 1 5 taD2 ^ tan 16 ' where or == £ (a -f /3 + y) • =£(180°-,4+180 o -.B+180 o -C) = 270°- l(A+B + C). What we have done amounts to interchanging the corresponding Greek and Roman letters. * it may be shown that S is the supplement of the diameter of the circumscribed circle. OBLIQUE SPHERICAL TRIANGLES 215 14. Napier's analogies. Dividing (20) by (23), we get tan \a _ I sin s sin (s — a) I sin s sin (s — b) ~ N sin (s — c) sin (s — a) or, '< sin s sin (s — a) sin(s — b) sin (s -o) " sin s sin (s — a) sin(s — 6) sin (s — c) sin s sin (s — b ) sin J a cos I- a cos £ j3 \ sin (s — c) sin (s — a) sin £ a cos J /? _ sin (s — a) cos £ a sin £ fi sin (s — i) By composition and division, in pro-portion, sin \ a cos J /? + cos ^ « sin J fi _ sin (,s — «) + sin (s — &) sin J a cos £ /3 — cos \ a sin J /3 sin (s — a) — sin (s — b) Erom (40), (41), p. 63, and (66), p. 74, Granville's Plane Trigo- nometry, the left-hand member equals smQa+J^ sin(i«- J/3)' and the right-hand member sin (s — a) + sin(s — &) _ tan J [s — q. + (s — 5)] _ tan Jc # sin(s — a) — sin (s — 6) tan J [s — a —(s — 5)] tan J (b — a) Equating these restilts,we get, noting that tan ^(b— a)=— tan£(a— 5), sin J (a + /J) _ tan ^c sin J (a — /?) tan J (a — 6) ' <»> «-»<— >~=i&g-K In the same manner we may get the two similar formulas for tan J (b — c) and tan \ (c — a). Multiplying (20) and (23), we get I sin s sin (s — a) sin s sin (s — b) tan 1 a tan 18= "V^-7 , x \ . — ^ AJ^ r-^— 7 — ^> * ¥ r \ sin (s — J) sin (s — c) N sin (s — c) sin (s — a) sin 1 a sin J # sin s cos £ a cos J /? sin (s — c) By composition and division, in proportion, cos J « cos J 8 — sin J or sin J 8 sin (s — c) — sin s cos £ a cos £ /3 + sin J a sin J S sin (s — c) + sin s • Fiir»-a + »-6=2«-o-J=o + 6 + c — o-6 = f, and s-a-$ + b=b — a. 216 SPHERICAL TRIGONOMETRY From (42), (43), p. 63, and (66), p. 74, Granville's Plane Trigo- nometry, the left-hand member equals cos (\ a + % /?) . cos (i a - i /8) ' and the right-hand member sin (.i — c) — sin s _ tan ^ (s — e — s) _ tan \ (— c) * sin (s — c) + sin s tan ^(s — c + s) tan £ (a. + b) Equating these results, we get, noting that tan £ (— c) = — tan ^ ?, cos £ (a + j3) _ tan ^ c cos i (« — y8) tan J (a + 6) ' (45) tani(a + ft) = - C ° Si(a ~ ^ tan j c. K J K ' cos J (a + 0) 2 In the same manner we may get the two similar formulas for tan \ (b + e) and tan £ (c + »)• By making use of the Principle of Duality on p. 208,we get at once from formulas (44) and (45), (46) u»l(.-/0— £}g^t»l y , (47) tan|(a + ^) = -^^ta„| y . By changing the letters in cyclic order we may at once write down the corresponding formulas for tan ^ (fi — y), tan ^ (y — a), tan £ (B+y), and tan i (y + «)■ The relations derived in this section are known as Napier's analogies. Since cos £(«-&) and tan h 7 = tan i ( 180 ° - c ) = tan ( 90 ° ~h c ) = cot J C are always positive, it follows from (47) that cos \(a + b) and tan \ (a + B) always have opposite signs ; or, since tan £(« + /3) = tan ^(180° -4 + 180° -.8) = tan £[360° -(,4 + £)] = tan [180°- £ {A + B)] = — tan£ (A + B), we may say that cos £ (a + b) and tan |(^[ + B) always have the same sign. Hence we have the Theorem. In a spherical triangle the sum of any two sides is less than, greater than, or equal to 180°, according as the sum of the cor- responding opposite angles is less than, greater than, or equal to 180°. 15. Solution of oblique spherical triangles. We shall now take up the numerical solution of oblique spherical triangles. There are three cases to consider with two subdivisions under each case. * For s-c-s = -c, and s — c + s = 2s-c = a + b + c-c=a + b. OBLIQUE SPHERICAL TRIANGLES 217 Case I. Case II. Case III. 16. Case I. namely, (27) (28) (29) (30) to find a, /?, (a) Given the three sides. (b) Given the three angles. (a) Given two sides and their included angle. (b) Given two angles and their included side. (a) Given two sides and the angle opposite one of them. (b) Given two angles and the side opposite one of them. (a) Given the three sides. Use formulas from p. 213, tan^rf jsin (s — a) sin (s — b) sin (s — c) \ sins tan I a = tan ^ d tan $ fi = tan ^ y = sin (s — a) tan ^-<2 sin (s — b) tan £ d sin (s — c) tan I d y, and therefore A, B, C, and check by the law of sines, sin a sin b sin c sin A sinB sin C Ex. 1. Given a = 60°, b = 137° 20', c = 116° ; find A, B, C. Solution. a = 60° 6 = 137° 20' c = 116° 2S: 313° 20' 8 = 156° 40'. s - a = 96° 40'. s - 6 = 19° 20'. s - c = 40° 40'. To find A use (28) logsin(s-a) = 9.9971 log tan £ d = 9^ To find log tan ^ d use (27) log sin (s- a) = 9.9971 log sin (s- 6)= 9.5199 log sin (s-c) = 9.8140 29.3310 log sin s = 9.5978 2| 19.7332 log tan ^ d = 9. log tan ^.a = 0.1305 i or = 53° 29'. or = 106° 58'. . ..4=180°-106°58'=73°2'. Check: log sin a = 9.9375 log sin^. = 9.9807 To find Buse (29) log sin (s- 6) = 9.5199 ; tan \ d = 9J log tan ^/3 = 9.6533 \ /3 = 24° 14'. = 48° 28'. . B=180°-48°28'=131°32'. log sin 6 = 9.8311 log sin B = 9.8743 9.9568 To find C use (30) log sin (s -c) = 9.8140 log tan £• d = 9.8666 log tan J y = 9.9474 £ Y = 41° 32'. 7 = 83° 4'. •. C=180°-83°4'=96°56'. log sin c = 9.9537 log sin C = 9J. This checks up closer than is to be expected in general. There may be a variation of at most two units in the last figure when the work is accurate. 218 SPHERICAL TRIGONOMETRY EXAMPLES Solve the following oblique spherical triangles : No. Given Parts Required Parts 1 a = 38° 6=51° c = 42° ^.= 51° 58' B = 83° 54' C = 58° 53' 2 a = 101° 6 = 49° c = 60° A=Ui° 32' B = 27° 52' C = 32° 28' 3 a = 61° 6 = 39° c = 92° ^1=35° 32' B = 24° 42' C = 138° 24' 4 a = 62° 20' b = 54° 10' c = 97° 50' ^1=47° 22' B = 42° 20' C = 124° 38' 5 a = 58° 6 = 80° c = 96° ^4 = 55° 58' B=74°14' C = 103° 36' 6 a = 46° 30' b = 62° 40' c = 83° 20' yl=43°58' 5 = 58° 14' C=108°6' 7 a = 71° 15' 6 = 39° 10' c = 40° 35' A =130° 36' B = 30°26' C = 31°26' 8 a = 47° 30' 6 = 55° 40' c = 60° 10' .4=56° 32' B = 69° 7' C = 78°58' 9 a = 43° 30' 6 = 72° 24' c = 87° 50' 4 = 41° 27' .8= 66° 26' C = 106° 3' 10 a =110° 40' b = 45° 10' c = 73° 30' .4=144° 27' B = 26° 9' C = 36°35' 17. Case I. (6) Given the three angles. Use formulas from p. 214, namely,* = l sin(,r-« ) sin(o--flsin ifL - x ) > > sin <r sin (a-. — a) (40) (41) (42) (43) tan \a = tan £8 sin (er — )8) tan *b— — -^ — — r^ > tan £8 tan^c _ sin (o- — y) tan £ 8 io ,/stmZ a, b, c ; and check by the law of sines, sin a _ sin b _ sin c sin A sin B sinC Ex. 1. Given A = 70°, B = 131° 10', C = 94° 50'; find a, 6, c. Solution. Here we use the supplements of the angles. To find log tan \ S use (40) a = 180°- A = 110° /3 = 180°-JB = 48° 50' 7 = 180°- C = 85° 10' 2 <r = 244° <r = 122°. a- a= 12°. <r - /3 = 73° 10'. <r _ 7 = 36° 50'. log sin (<r - a) - 9.3179 log sin (<r - (8) = 9.9810 log sin (<r - y) = 9.7778 29.0767 log sin <r = 9.9284 2 | 19 1483 log tan £ <5= 9.5742 * These formulas may be -written down at once from those used in Case I, (a), p. 217, by simply interchanging the corresponding Greek and Roman letters. OBLIQUE SPHERICAL TRIANGLES 219 To find a use (41) log sin (a- - a) = 9.3179 log tan £ S = 9.5742 logtan|a = 9.7437 \ a = 29°. .-. a = 58°. To find 6 use (42) log sin (<r-0) = 9.9810 log tan % 8 = 9.5742 log tan£& = 0.4068 %b = 68° 36'. 6 = 137° 12'. Check : log sin a = 9.9284 log sin 4 = 9.9730 9.9554 log sin 6 = 9.8321 log sin£ = 9.8767 9.9554 To find c use (43) log sin (o-- 7) = 9.7778 log tan \ 5 = 9.5742 logtan£c = 0.2036 £c = 57°58'. c = 115° 56'. log sin c = 9.9539 log sin C = 9. 9985 9.9554 EXAMPLES Solve the following oblique spherical triangles : No. Given Parts Required Parts 1 4=75° £ = 82° C = 61° a = 67° 52' 6 = 71° 44' c = 57° 2 4=120° £ = 130° C = 80° a = 144° 10' .6 =148° 49' c = 41° 44' 3 4 = 91° 10' £ = 85° 40' C = 72° 30' a = 89° 51' 6 = 85° 49' c = 72° 32' 4 4 = 138° 16' £ = 31° 11' C = 35°53' a = 100° 5' 6 = 49° 59' c = 60° 6' 5 A =78° 40' £ = 63° 50' C = 46°20' a = 39° 30' 6 = 35° 36' c = 27° 59' 6 4=121° £ = 102° C = 68° a = 130° 50' 6 = 120° 18' c = 54° 56' 7 4=130° £ = 110° C = 80° a = 139° 21' 6 = 126° 58' c = 56° 52' 8 4 = 28° £ = 92° C = 85°26' a = 27° 56' 6 = 85° 40' c = 84° 2' 9 4= 59° 18' £ = 108° C = 76°22' a = 61° 44' 6= 103° 4' c = 84° 32' 10 4=100° £=100° C = 50° a =112° 14' 6 = 112° 14' c = 46°4' 18. Case II. (a) Given two sides and their included angle, as a, b, C. Use formulas on p. 216, namely, (46) (47) , / „x sin 4- (a — b) , tani(a: — /?) = :— 7-7 rf tan Ay, ^ v ^ y sin£(a + &) 2 " cos 1 (a — b) ^ tan £(« + £) = y) — -rftaniy, ^ v ^' cos £(« + £) ' to ^jwZ a and (i and therefore A and B ; and from p. 215 use (44) solved for tan \ e, namely, sin j(a + /3)tan j- (a — b) (44) tan i c = sin£(a — /3) to find c. Cheek by the law of sines Ex. 1. Given a = 64° 24', 6 = 42° 30', C = 58° 40' ; find 4, £, c. Solution. 7 = 180° - C = 121° 20'. .-. i 7 = 60° 40'. a= 64° 24' 6 = 42° 30' a + b = 106° 54' .-. \ (a + 6) = 53° 27'. a = 64° 24' 5 = 42° 30' a - 6 = 21° 54' .-. i (a -6) = 10° 57'. 220 SPHERICAL TRIGONOMETRY To find i (a- p) use (46) log sin b (a - b) = 9. 2786 log tan^7 = 0.2503 9.5289 log sin -^ (a + 6) = 9.9049 log tan-£(« - p) = 9.6240 (n) .-. £(a-/3) = -22°49'.* To find A and B b(a + p)= 108° 49' i(a-p)=- 22° 49' Adding, a = 86° Subtracting, p = 131° 38'. .-. A = 180° - a = 94°. 5 = 180° - = 48° 22'. ToJtnd^(a: + /3) use (47) log cos £ (a- 6) = 9.9920 log tan 4- 7= 0.2503 10.2423 log cos£ (a + 6) = 9.7749 log tan b(a + p)= 0.4674 (n) 180° -•£(« + £) = 71° ll'.t .-. | (a +0) = 108° 49'. To find c use (44) logsin£(a + jS)= 9.9761 log tan | (a - 6) = 9.2867 19.2628 log sin £ (a- - /3) = 9.5886 (n) logtan£c = 9.6742 J £c = 25°17'. . . c = 50° 34'. C%ec& : log sin a = 9.9551 log sin A = 9S log sin 6 : log sin B -. 9.9562 = 9.8297 : 9.8735 9.9562 log sin c = 9.8878 log sin C = 9.9315 9.9563 If o only is wanted, we may find it from the law of cosines, (14), p. 209, without previously determining A and B. But this formula is not well adapted to logarithmic calculations. Another method is illustrated below, which depends on the solu- tion of right spherical triangles, and hence requires only those formulas which follow from applying Napier's rules of circular parts, p. 200. Through B draw an arc of a great circle per- pendicular to A C, intersecting A C at D. Let BD=p, CD = m, AD = n. Applying Rule I, p. 200, to the right spherical triangle BCD, we have cos C = tan m cot a, or, (.4) tan m = tan a cos C. Applying Rule II, p. 200, to BCD, cos a = cos to cos^, or, (B) cosp = cos a sec to. * Since tan £ (a -8) is negative, £(ar — 0) may be an angle in the second or fourth quad- rant. But a > b, therefore A> B and a < 8, since a and 8 are the supplements of A and B. Hence \{a — 8) must be a negative angle numerically less than 90°. t Here £(a + 8) must be a positive angle less than 180°. Since tan $ (a +,8) is negative, $(« + 8) must lie in the second quadrant, and we get its supplement from the table. } tan J c is positive, since sin j (a- a) is negative and there is a minus sign before the fraction. OBLIQUE SPHERICAL TRIANGLES Applying the same rule to ABB, cos c = cos n cosjj, or, (C) cosp = cos o sec n. Equating (B) and (C), cos o sec n = cos a sec m, or, cos c = cos a sec »i cos n. But ?t = J — m ; therefore (-D) cos c = cos a. sec m cos (i — m). i Now c may be computed from (4) and (Z>), namely, (48) tan m = tan a cos C, 221 (49) cos c _ cos a cos (6 — m) cos m Ex. 2. Given a = 98°, 6 = 80°, C = 110° ; find c. Solution. Apply the method just explained. To find b — m use (48) log tan a = 0.8522 (n) log cos C = 9.5341 (n) log tan m = 0.3863 m = 67° 40'. .-. b-m = 12° 20'. To find c use (49) log cos a = 9.1436 (n) ;cos(6-ro) = 9.98 99 19.1335 log cos m = 9.5798 log cos c = 9. 5537 (to) 180° - c = 69° 2'. c = 110° 58'. EXAMPLES Solve the following oblique spherical triangles : No. Given Parts Required Parts 1 6 = 137° 20' c = 116° A = 70° £ = 131° 17' C = 94°48' a = 57° 57' 2 a = 72° 6 = 47° C = 33° A = 121° 33' £ = 40° 57' c = 37°26' 3 a = 98° c = 60° £ = 110° ^4=87° C=60°51' b = 111° 17' 4 6 = 120° 20' c = 70°40' ^4 = 50° B - 134° 57' C = 50° 41' a = 69° 9' 5 a = 125° 10' b = 153° 50' C = 140° 20' A = 147° 29' £ = 163° 9' c=76°8' 6 a = 93° 20' 6 = 56° 30' C = 74° 40' .4 = 101° 24' £ = 54° 58' c=79°10' 7 6 = 76° 30' c = 47° 20' A = 92° 30' £ = 78° 21' C = 47°47' a = 82° 42' 8 c = 40°20' a = 100° 30' £ = 46° 40' A = 131° 29' C = 29° 33' 6 = 72° 40' 9 6 = 76° 36' c = 110° 26' ^1 = 46° 50' £ = 57° 38' C = 125° 32' a = 57° 8' 10 a = 84° 23' b = 124° 48' C = 62° ^1=68° 27' £ = 129° 53' c= 70° 52' 222 SPHERICAL TRIGONOMETRY 19. Case II. (b) Given two angles and their included side, as A, B, c. Use formulas * o% pp. 215, 216, namely, (44) tan Ha - &) = - ^ ti" T ff tan £ c > (45) tani(a + b) = - °° S \ j" ~ ff tan£ e , to ./iwrf a and b ; and from p. 216, use (46) solved for tan £ y, namely, sin $(a + b) tan ■$■ (a — /?) (46) tan £ y ■■ sin £ (a — 6) to ./md y and therefore C. Check by the law of sines. Ex. 1. Given c = 116°, A = 70°, B = 131° 20' ; find a, 6, C. Solution, a = 180° - ^ = 110°, and /3 = 180° - 5 = 48° 40'. a = 110° p = 48° 40' a + /3 = 158°40' .-. £(a + /3) = 79°20'. To find i (a - 6) use (44) logsin£(a-/3) = 9.7076 log tan ^-c = 0.2042 9.9118 logsin £(<* + P) = 9.9924 log tan \ (a - b) = 9.9194 (n) .-. £(a-6) = -39°43'.t a = 110° ft = 48° 40' a-p = 61°20' £(<*-/3) = 30°40'. c = 116°. £c = 58°. To find a and b To find i (a + 6) use (45) log cos i(a~ p)= 9.9346 log tan £ c = 0.2042 10.1388 log cos -J- {a + p) = _9 1 2674 log tan £ (a + 6) = 0.8714 (n) 180° -£(a + b) = 82° 21'. .. |(a + &) = 97°39'. To find C use (46) log sin £ (a + 6) = 9.9961 log tan £(a - /3) = 9.7730 19.7691 log sin £ (o - 6) = 9.8055 (n) log tan £7 = 9.9636 £7 = 42° 36'. 7 = 85° 12'. .-. = 180° - 7 = 94° 48'. Cftecfc : log sin a = 9.9281 log sin 6 = 9.8308 log sin c = 9.9537 logsin^. = 9.9730 log sin B = 9.8756 logsinC = 9.E 9.9651 % (a + 6) = 97° 39' | (a -ft) = - 39° 43' Adding, Subtracting, a = 67° 56' 6 = 137° 22'. 9.9562 9.9552 * Same as those used in Case II, (a), p. 219, with Greek and Roman letters interchanged, t Since A < B it follows that a < b, and } (a - 6) is negative. OBLIQUE SPHERICAL TRIANGLES 223 If C only is wanted, we can calculate it without previously determining a and b, by dividing the given triangle into two right spherical triangles, as was illustrated on p. 220. Through B draw an are of a great circle perpendicular to AC, intersecting AC at D. Let BD=p, angle ABB = x, angle CBD = y. Applying Rule I of Napier's rules, p. 200, to the right spher- ical triangle ABD, we have cos o = cot x cot A, or, (A) cot x = tan A cos c. Applying Rule II, p. 200, to ABD, we have cos A = oosp sin x, or, (B) cosp = cos A esc x. Applying the same rule to CBD, cos C = cosp sin y, or, (C) cosp = cos C esc y. Equating (B) and (C), cos C esc y = cos A esc a;, or, cos C = cos vl esc x sin ?/. But y = B — x ; therefore (D) cos C = cos A esc a; sin (B — x). Now C may be computed from (A) and (D), namely, (50) cot x = tan .4 cos c. cos 4 sin (B — x) (51) cos C = ■ sin x Ex. 2. Given ^. = 35° 46', B = 115° 9', c = 51° 2' ; find 0. Solution. Apply the method just explained. To find B- x use (50) log tan .4 = 9.8575 log cose = 9. 7986 log cot a; = 9.6561 x = 65° 38'. .-. B - x = 49° 31'. To find C use (51) log cos A = 9.9093 log sin (B - x) = 9.8811 19.7904 log sin x = 9. 9595 log cos C= 9.8309 C = 47°21'. 224 SPHERICAL TEIGONOMETEY EXAMPLES Solve the following oblique spherical triangles : No. Given Parts Required Parts 1 .4=67° 30' S = 46° 50' c=74°20' a = 63° 15' 6 = 43° 53' C = 95°l' 2 B = 98° 30' (7 = 67° 20' a = 60° 40' 6 = 86° 40' c = 68°40' ^=59° 44' 3 C=110° ^1=94° 6 = 44° a = 1 14° 10' c = 120° 46' B = 49° 34' 4 C = 70°20' B = 43° 50' o = 50° 46' 6 = 32° 59' c = 47°45' ^. = 80° 14' 5 A =78° B = 41° c = 108° a = 95°38' 6 = 41°52' C = 110°49' 6 £ = 135° C = 50° a = 70° 20' 6 = 120° 16' c = 69° 20' ^.= 50° 26' 7 A =31° 40' C=122°20' 6 = 40° 40' a = 34° 3' c=64°19' .8 = 37° 40' 8 A = 108° 12' £ = 145° 46' c =126° 32' a = 69° 5' 6 =146° 25' C = 125°12' 9 A = 130° 36' B = 30° 26' c = 40° 35' a =71° 15' 6 = 39° 10' C = 31°26' 10 .4=51° 58' E = 83° 54' c = 42° a = 38° 6=51° C=58°53' 20. Case III. (a) Given two sides and the angle opposite one of them, as a, b, B (ambiguous case *). From the law of sines, p. 207, we get sin a sin B (11) sin 4 : sin b which gives A^. To find C we use, from p. 216, formula (46), solved for tan \ y, namely, (46) tan | y = — sin|(a + ft)tan§(q-ff) sin I (a — 6) To find c, solve (44), p. 215, for tan £ c, namely, (44) Chech by the law of sines. tnn i r _ sinj(a + fltan|( a -&) sin§(a-£) Ex. 1. Given a = 58°, 6 = 137° 20', B = 131° 20'; find A, C, c. Solution. To find A use (11) log sin a = 9.9284 log sin B = 9.8756 19.8040 log sin 6 = 9.8311 log sin A= 9.9729 . . A 1 = 69° 58', A 2 = 180° - Ai = 110° 2'. a =58° 6 = 137° 20' a =58° 6= 137° 20' o + 6=195°20' o-6=-79°20' \ (o+ 6) = 97° 40'. \ (a -6) =-39° 40'. = 180° -.8 = 48° 40'. Since a<b and both Ai and A 2 are<JB, it follows that we have two solutions. * Just as in the corresponding case in the solution of plane oblique triangles (Granville's Plane Trigonometry, pp. 105 t 161), there may be two solutions, one solution, or no solution, depending on the given data. t Since the angle A is here determined from its sine, it is necessary to consider both of the values found. If a>b then A>B; and if a<b then^<£. Hence [next page] OBLIQUE SPHEEICAL TRIANGLES 225 First solution, a x = 180° - A-y = 110° 2'. a x = 110° 2' P= 48° 40' a x + (8 = 158° 42' £(ai + |8)=79°21'. To^nd d use (46) log sin £ (a + 6) = 9.9961 log tan \ {a x -p) = 9.7733 19.7694 log sin t(a-b)= 9.8050 (n) log tan ^7i = 9.9644 £71 = 42° 39'. 71 = 85° 18'. .-. d = 180° - 7l = 94° 42'. Cftecfc : log sin a = 9. 9284 log sin 6 log sin A x — 9.9729 log sin B 9.9555 <*! = 110° 2' j3= 48° 40' ai-/3= 61° 22' £(ai-/?)=30°41'. To find ci use (44) log sin £ (orj + p) = 9.9924 log tan I (a - 6) = 9.9187 (n) 19.9111 log sin £ (a! - |3) = 9.7078 log tan £c x = 10.2033 •Jci = 57°57'- .-. Ci = 115° 54'. = 9.8311 = 9.8756 9.9555 log sin ci = 9.9541 log sin Ci = 9.9985 9.9556 Second solution. a 2 = 180° — X 2 = 69 e 58'. or 2 = 69° 58' |8= 48° 40' Q-2 + P = 118° 38' £(a 2 + ,3) = 59°19'. To find C 2 wse (46) logsin£(a + &) = 9.9961 log tan \ (or 2 - p) = 9.2743 19.2704 log sin £ (a - 6) = 9.8050 (m) log tan £72= 9.4654 ■£-72 = 16° 17'. 7 2 = 32° 34'. .-. C 2 = 180° - 72 = 147° 26'. Check : log sin a = 9.9284 log sin 6 log sin _4 2 = 9.9729 log sin B 9.9555 a 2 = 69° 58' p = 48° 40' a 2 - p = 21° 18' £(a 2 -/3) = 10°39'. To find C\ use (44) logsin£(a- 2 + (3) = 9.9345 log tan £ (a - 6) = 9.9187 (n) 19.8532 log sin £ (ar 2 - p) - 9.2667 log tan \ c 2 = 10.5865 £c 2 = 75°28'. . . c 2 = 150° 56'. : 9. 831 1 log sin c 2 = 9. 6865 : 9.8756 log sin C 2 = 9. 7310 9.9555 9.9555 If the side c or the angle C is wanted without first calculating the value of A, we may resolve the given triangle into two right triangles and then apply Napier's rules, as was illustrated under Cases II, (a), and II, (b), pp. 220, 223. Theorem. Only those values of A should be retained which, are greater or less than B according as a is greater or less than b. If log sin A = a positive number, there will be no solution. 226 SPHERICAL TRIGONOMETRY EXAMPLES Solve the following oblique spherical triangles : No. Given Parts Required Pakts 1 a =43° 20' 6=48° 30' 4 = 58° 40' B!=68°46' C 1= 70°46' £ 2 =111°14' C 2 = 14°29' ci=49°16' c 2 = ll°36' 2 a =56° 40' 6=30° 50' 4 = 103° 40' £=36° 35' C=52° c=42°39' 3 a =30° 20' 6=46° 30' A =36° 40' £i = 59°4' Ci = 97°39' B2 = 120°56' C 2 =28°5' C! = 56°57' c 2 =23°28' 4 6=99° 40' c=64°20' £=95° 40' C=65°30' 4 = 97°20' a=100°45' 5 a =40° 6=118° 20' 4 = 29° 40' £ 1= 42°40' Ci = 159°54' £ 2 = 137°20' C 2 =50°21' Ci = 153°30' c 2 =90°10' 6 a = 115°20' c = 146°20' C=141°10' Impossible 7 a=109°20' c=82° A = 107° 40' C=90° £=113°35' 6=114° 50' 8 6=108° 30' c=40°50' C=39°50' £i=68°18' 4i = 132°34' £ 2 = 111°42'4 2 = 77°5' ai=131°16' o 2 =95° 50' 9 a =162° 20' 6 = 15° 40' £=125° Impossible 10 a =65° c = 138°10'^=42°30' 0=146° 38' £=55° 1' 6=96° 34' 21. Case III. (6) Given two angles and the side opposite one' of them, as A, B, b (ambiguous case *). From the law of sines, p. 207, we get BVO.A sin b (11) sin a = sinB which gives a.| To find c we use, from p. 215, the formula % (44), solved for tan £ o, namely, tan 1 ^ Bin§(a + ff)tan§(q-&) 2 sm\{a-P) (44) To find 0, solve (46), p. 216, for tan\ y, namely, , , , sin§(a + &)tan§(a — 8) (46) tan|y = aV TV ^ ^. v J * * sin|(a — b) Check by the law of sines. * Just as in Case II, (&), we may have two solutions, one solution, or no solution, depending on the given data. t Since the side is here determined from its sine, it is necessary to examine both of the values found. If A > B then a > b ; and if A < B then a < b. Hence we have the Theorem. Only those values of a should be retained which are greater or less than b according as A is greater or less than B. If log sin a = a positive number, there will be no solution. % Same as those used in Case III, (a), p. 224, when the Greek and Roman letters are interchanged. OBLIQUE SPHERICAL TRIANGLES 227 Ex. 1. Given A = 110°, B = 131° 20', 6 = 137° 20' ; find a, c, C. Solution, a = 180° -A = 70°, and /3 = 180°-B = 48° 40'. To find a use (11) log sin A= 9.9730 log sin 6 = 9.8311 19.8041 log sin B= 9.8756 log sin a = 9.9285 .-. ai = 58° 1', a! = 121° 59'. . = 180° ■ a = 70° ,8= 48° 40' -/3= " a = 10° j3 = 48°40' First solution. (i!= 58° 1' 6 = 137° 20' a! + 6 = 195° 21' |(a 1 + 6) = 97°41'. To find Ci use (44) log sin £ (a + j3) = 9. 9346 log tan \ (en - 6) = 9.9187 (n) 19.8533 log sin \ (a - |3) = 9. 2674 log tan £ci = 10. 5859 £ci = 75°27'. .-. ci = 150° 54'. Cheek: log sin ai = 9.9285 log sin 6 = 9.8311 log sin ci = 9.6 log sin A = 9.9730 log sin B = 9.8756 log sin Ci = 9. 7314 9.9555 a + /3 = 118°40' <x-/3 = 21°20' b (a + 0) = 59° 20'. \ {a - p) = 10° 40'. Since A<B and both ai and a 2 are < 6, it follows that we have two solutions. ai= 58° V 6 = 137° 20' ai - 6 = - 79° 19" £(ai-6) = -39°40'. Toured Ci use (46) logsin£(ai + 6) = 9.9961 log tan % (a - j3) = 9.2750 19.2711 log sin £ (ai - 6) = 9.8050 (n) log tan £ 7i= 9.4661 £71 = 16° 18'. 71 = 32° 36'. .-. Ci = 180° -71 = 147° 24'. 9.9555 9.9555 Second solution. This gives c 2 = 64° 8', and C 2 = 85° 18'. Remembering that a 2 = 121° 59', we may now check the second solution. Check : log sin a 2 = 9. 9285 log sin 6 = 9. 8311 log sin c 2 = 9. 9542 log sin A = 9.9730 log sin B = 9.8756 log sin C 2 = 9.9985 9.9555 9.9555 9.9557 Hence the two solutions are ai = 58° 1' ci = 150° 54' Ci = 147° 23', and a 2 = 121°59' c 2 = 64°8' C 2 = 85° 18'. If the angle C or the side is wanted without first computing a, we may resolve the given triangle into two right triangles and then apply Napier's rules, as was illustrated under Cases II, (a), and II, (b), pp. 220, 223. 228 SPHERICAL TRIGONOMETRY EXAMPLES Solve the following oblique spherical triangles : No. Given Parts Required Parts 1 4 = 108° 40' .8=134° 20' a =145° 36' 6 = 154° 45' c=34°9' C=70°18' 2 £=116° C=80° c=84° 6 = 114° 50' 4 = 79° 20' a =82° 56' 3 4 = 132° £=140° 6=127° Oi=67° 24' d = 164°6' a 2 =112°36'C 2 = 128°21' Ci = 160°6' c 2 =103°2' 4 4=62° C=102° a =64° 30' c=90° £=63° 43' 6=66° 26' 5 4 = 133° 50' £=66° 30' a = 81° 10' Impossible 6 5=22° 20' C = 146°40' e = 138°20' 6=27° 22' 4 = 47° 21' a = 117°9' 7 4 = 61° 40' C=140°20' c=150°20' a 1= 43°3' £ 1 = 89°24' a 2 = 136°57'£ 2 =26°59' 6! = 129°8' 6 2 =20°36' 8 £=73° C=81°20' 6= 122° 40' Impossible 9 £=36° 20' C=46°30' 6=42° 12' 4 1 = 164°44' <i 1 = 162°88' 4 2 =119°17' a 2 = 81°17' C! = 124°41' c 2 =55°19' 10 4 = 110° 10' £=133° 18' a=147°6' 6=155° 5' c=32°59' C=70°16' 22. Length of an arc of a circle in linear units. From Geometry we know that the length of an are of a circle is to the circumference of the circle as the number of degrees in the arc is to 360. That is L:2itR::N: 360, or, VRN Is — ■ ' ' '-' * 180 L = length of arc, N = number of degrees in arc, R = length of radius. In case the length of the arc is given to find the number of degrees in it, we instead solve for N, giving (53) N = 180 L TTR Considering the earth as a sphere, an arc of one minute on a great circle is called a geographical mile or a nautical mile.* Hence there are 60 nautical miles in an arc of 1 degree, and 360 x 60 = 21,600 nautical miles in the circumference of a great circle of the earth. If we assume the radius of the earth to be 3960 statute miles, the length * In connection with a ship's rate of sailing a nautical mile is also called a knot. OBLIQUE SPHERICAL TRIANGLES 229 of a nautical mile (= 1 min. = 5 \ of a degree) in statute miles will be, from (52), 3.1416 x 3960 x A 11S . L- m = 1.15 mi. Ex. 1. Eind the length of an arc of 22° 30' in a circle of radius 4 in. Solution. Here N = 22° .30' = 22. 5°, and B = 4 in. Substituting in (52), L = 3 - 1416 x 4 x 22.5 = ^ .^ ^ 180 Ex. 2. A ship has sailed on a great circle for 5J hr. at the rate of 12 statute miles an hour. How many degrees are there in the arc passed over ? Solution. Here L = 5£ x 12 = 66 mi. , and B = 3960 mi. Substituting in (53), N = 18 ° X 66 = .955° = 57.3'. Ans. 23. Area of a spherical triangle. From Spherical Geometry we know- that the area of a spherical triangle is to the area of the sur- face of the sphere as the number of degrees in its spherical excess * is to 120. That is, Area of triangle : 4 ttR 2 :: E : 720, or, TTIPE (54) Area of spherical triangle : 180 In case the three angles of the triangle are not given, we should first find them by solving the triangle. Or, if the three sides of the tri- angle are given, we may find E directly by Lhuilier's formula, f namely, (55) tan \E = Vtan | s tan § (s — c )tan§(s — 6)tan|(s— c), where a, h, c denote the sides and s = \ (a + b + c). The area of a spherical polygon will evidently be the sum of the areas of the spherical triangles formed by drawing arcs of great cir- cles as diagonals of the polygon. Ex. 1. The angles of a, spherical triangle on a, sphere of 25-in. radius are A = 74° 40', B = 67° 30', C = 49° 50'. Eind the area of the triangle. Solution. Here E = (A + B+C)- 180° = 12°. c w + *• • /ka\ a 3.1416 x (25)2 x 12 19no . . Substituting in (54), Area = — = 130.9 sq. in. Ans. * The spherical excess (usually denoted by E) of a spherical triangle is the excess of the sum of the angles of the triangle over 180°. Thus, if A, B, and C are the angles of a spher- ical triangle, E=A + B+C-1W>. t Derived in more advanced treatises. 230 SPHERICAL TRIGONOMETRY EXAMPLES 1. Find the length of an arc of 5° 12' in a circle whose radius is 2 ft. 6 in. Ans. 2.72 in. 2. Find the length of an arc of 76° 30' in a circle whose radius is 10 yd. Ans. 13.17 yd. 3. How many degrees are there in a circular arc 15 in. long, if the radius is 6 in.? Ans. 143° 18'. 4. A ship sailed over an arc of 4 degrees on a great circle of the earth each day. At what rate was the ship sailing ? Ans. 11.515 mi. per hour. 5. Find the perimeter in inches of a spherical triangle of sides 48°, 126°, 80°, on a sphere of radius 25 in. Ans. 110.78 in. 6. The course of the boats in a yacht race was in the form of a triangle having sides of length 24 mi., 20 mi., 18 mi. If we assume that they sailed on arcs of great circles, how many minutes of arc did they describe ? Ans. 53.85 min. 7. The angles of a spherical triangle are A = 63°, B = 84° 21', C = 79°; the radius of the sphere is 10 in. What is the area of the triangle ? Ans. 80.88'sq. in. 8. The sides of a spherical triangle are a — 6.47 in. , 6 = 8.39 in. , c = 9.43 in. ; the radius of the sphere is 25 in. What is the area of the triangle ? Ans. 26.9 sq. in. Hint. Find E by using formula (55)' 9. In a spherical triangle A = 75° 16', B = 39° 20', c = 26 ft. ; the radius of the sphere is 14 ft. Find the area of the triangle. Ans. 158.45 sq. ft. 10. Two ships leave Boston at the same time. One sails east 441 mi. and the other 287 mi. E. 38° 21' N. the first day. If we assume that each ship sailed on an arc of a great circle, what is the area of the spherical triangle whose ver- tices are at Boston and at the positions of the ships at the end of the day ? Ans. 41,050 sq. mi. 11. A steamboat traveling at the rate of 15 knots per hour skirts the entire shore line of an island having the approximate shape of an equilateral triangle in 18 hr. What is the approximate area of the island ? Ans. 4651.1 sq. mi. 12. Find the areas of the following spherical triangles, having given (a) a = 47° 30', 6 = 55° 40', c = 60° 10'; B = 10 ft. Ans. 42. 96 sq. ft. (b) a = 43° 30', 6 = 72° 24', c = 87° 50'; R = 10 in. 59.21 sq. in. (c) A = 74° 40', B = 67° 30', C = 49° 50'; B = 100 yd. 2094 sq. yd. (d) A = 112° 30', B = 83° 40', C = 70° 10'; B = 25 cm. 941.4 sq. cm. (e) a = 64° 20', b = 42° 30', C = 50° 40'; B = 12 ft. 46.74 sq. ft. (f) C = 110°, A = 94°, 6 = 44° ; B = 40 rd. 2056.5 sq. rd (g) a = 43° 20', b = 48° 30', A = 58° 40'; B = 100 rd. 19.76 acres, (h) A = 108° 40', B = 134° 20', o = 145° 36'; B = 3960 mi. 36,466,667 sq. mi. CHAPTEE III APPLICATIONS OF SPHERICAL TRIGONOMETRY TO THE CELESTIAL AND TERRESTRIAL SPHERES 24. Geographical terms. In what follows we shall assume the earth to be a sphere of radius 3960 statute miles. The meridian of a place on the earth is that great circle of the earth which passes through the place and the north and south poles. (North pole) N W (West) S (South pole) Thus, in the figure representing the earth, NGS is the meridian of Greenwich, NBS is the meridian of Boston, and NCS is the merid- ian of Cape Town. The latitude of a place is the arc of the meridian of the place ex- tending from the equator to the place. Latitude is measured north or south of the equator from 0° to 90°. Thus, in the figure, the arc LB measures the north latitude of Boston, and the arc TC measures the south latitude of Cape Town. The longitude of a place is the arc of the equator extending from the zero meridian * to the meridian of the place. Longitude is * As in this case, the zero meridian, or reference meridian, is usually the meridian pass- ing through Greenwich, near London. The meridians of Washington and Paris are also used as reference meridians. 231 232 SPHERICAL TRIGONOMETRY measured east or west from the Greenwich meridian from 0° to 180°. Thus, in the figure, the are MT measures the east longitude of Cape Town, while the arc ML measures the west longitude of Boston. Since the arcs MT and ML are the measures of the angles MNT and MNL respectively, it is evident that we can also define the longitude of a place as the angle between the reference meridian and the meridian of the place. Thus, in the figure, the angle MNT is the east longitude of Cape Town, while the angle MNL is the west longitude of Boston. The bearing of one place from a second place is the angle between the arc of a great circle drawn from the second place to the first place, and the meridian of the second place. Thus, in the figure, the bearing of Cape Town from Boston is measured by the angle CBN or the angle CBL, while the bearing of Boston from Cape Town is measured by the angle NCB or the angle SCB.* 25. Distances between points on the surface of the earth. Since we know from Geometry that the shortest distance on the surface of a (_~North pole) N (South pole) sphere between any two points on that surface is the arc, not greater than a semicircumference, of the great circle that joins them, it is evident that the shortest distance between two places on the earth is measured in the same way. Thus, in the figure, the shortest * The bearing or course of a ship at any point is the angle the path of the ship makes with the meridian at that point. APPLICATIONS OF SPHERICAL TRIGONOMETRY 233 distance between Boston and Cape Town is measured on the arc BC of a great circle. We observe that this arc BC is one side of a spherical triangle of which the two other sides are the arcs BN and CN. Since arc BN = 90°- arc LB = 90°- north latitude of Boston, arc CN = 90° + arc TC = 90° + south latitude of Cape Town, and angle BNC = angle MNL + angle MNT = west longitude of Boston + east longitude of Cape Town = difference in longitude of Boston and Cape Town, it is evident that if we know the latitudes and longitudes of Boston and Cape Town, we have all the data necessary for determining two sides and the included angle of the triangle BNC. The third side BC, which is the shortest distance between Boston and Cape Town, may then be found as in Case II, (a), p. 219. In what follows, north latitude will be given the sign + and south latitude the sign — . Rule for finding the shortest distance between two points on the earth and the bearing of each from the other, the latitude and longitude of each point being given. First step. Subtract the latitude of each place from 90°.* The results will be the two sides of a spherical triangle. Second step. Find the difference of longitude of the two places by subtracting the lesser longitude from the greater if both are E. or both are W., but add the two if one is E. and the other is W. This gives the included angle of the triangle.^ Third step. Solving the triangle by Case II, (a), p. 219, the third side gives the shortest distance between the two points in degrees of arc,% and the angles give the bearings. * Note that this is algebraic subtraction. Thus, if the two latitudes were 25° N. and 42° S., we would get as the two sides of the triangle, 90" -25° =65° and 90° -(-42°) =90° + 42° =132°. t If the difference of longitude found is greater than 180°, we should subtract it from 360° and use the remainder as the included angle. t The number of minutes in this arc will be the distance between the two places in geo- graphical (nautical) miles. The distance between the two places in statute miles is given by the formula 3.1416 x 3960 x iv" Mi= ' I 180 where N= the number of degrees in the arc. 234 SPHERICAL TRIGONOMETRY Ex. 1. Find the shortest distance along the earth's surface between Boston (lat. 42°21'N., long. 71° 4' W.) and Cape Town (lat. 33° 56' S., long. 18° 26' E.), and the bearing of each city from the other. Solution. Draw a spherical triangle in agreement with the figure on p. 232. First step. c = 90° -42° 21' = 47° 39', 6 = 90° - (- 33° 56') = 123° 56'. Second step. N = 71° 4' + 18° 26' = 89° 30' = difference in long. Third step. Solving the triangle by Case II, (a), p. 219, we get n = 68° 14' = 68.23° = 4094 nautical miles, = 52° 43', and JB = 116°43'. Hence a ship sailing from Boston to Cape Town on the arc of a great circle sets out from Boston on a course S. 63° 17' E. and approaches Cape Town on a course S. 52° 43' E.* EXAMPLES 1. Find the shortest distance between Baltimore (lat. 39° 17' N., long. 76° 37' W.) and Cape Town (lat. 33° 56' S., long. 18° 26' E.), and the bearing of each from the other. Ans. Distance = 180° — 65° 48' = 6852 nautical miles, S. 64° 58' E. = bearing of Cape Town from Baltimore, N. 57° 42' W. = bearing of Baltimore from Cape Town. 2. What is the distance from New York (lat. 40°43'N., long. 74° W.) to Liverpool (lat. 53° 24' N., long. 3° 4' W.)? Find the bearing of each place from the other. In what latitude will a steamer sailing on a great circle from New York to Liverpool cross the meridian of 50° W. , and what will be her course at that point 1 Ans. Distance = 47° 50' = 2870 nautical miles, N. 75° 7' W. = bearing of New York from Liverpool, N. 49° 29' E. = bearing of Liverpool from New York. Lat. 51° 13' N., with course N. 66° 54' E. 3. Find the shortest distance between the following places in geographical miles : (a) New York (lat. 40° 43' N. , long. 74° W. ) and San Francisco (lat. 37° 48' N. , long. 122° 28' W.). Ans. 2230. (b) Sandy Hook (lat. 40° 28' N. , long. 74° 1' W. ) and Madeira (lat. 32° 28' N. , long. 16°55'W.). Ans. 2749. ' (c) San Francisco (lat. 37° 48' N., long. 122° 28' W.) and Batavia (lat. 6° 9' S., long. 106° 53' E.). Ans. 7516. (d) San Francisco (lat. 37° 48' N., long. 122° 28' W.) and Valparaiso (lat. 33° 2' S., long. 71° 41' W.) Ans. 5109. * A ship that sails on a great circle (except on the equator or a meridian) must be con- tinually changing her course. If the ship in the above example keeps constantly on the course S. 63° 17' E., she will never reach Cape Town. APPLICATIONS OF SPHERICAL TRIGONOMETRY 235 4. Find the shortest distance in statute miles (taking diameter of earth as 7912 mi.) between Boston (lat. 42° 21' N., long. 71° 4' W.) and Greenwich (lat. 51° 29' N.), and the bearing of each place from the other. Ans. Distance = 3275 mi. , K 53° 7' E. = bearing of Greenwich from Boston, N. 71° 39' W. = bearing of Boston from Greenwich. 5. As in last example, find the shortest distance between and bearings for Calcutta (lat. 22° 33' N, long. 88° 19' E.) and Valparaiso (lat. 33° 2' S., long. 71°42'"W.). Ans. Distance = 11,012.5 mi., S. 64° 20.5' E. = bearing of Calcutta from Valparaiso, S. 54° 54. 5' W. = bearing of Valparaiso from Calcutta. 6. Find the shortest distance in statute miles from Oberlin (long. 82° 14' W.) to New Haven (long. 72° 55' W.), the latitude of each place being 41° 17' N. Ans. 483.3 mi. 7. From a point whose latitude is 17° N. and longitude 130° W. a ship sailed an arc of a great circle over a distance of 4150 statute miles, starting S. 54°20' W. Find its latitude and longitude, if the length of 1° is 69£ statute miles. Ans. Lat. 19° 42' S., long. 178° 21' W. 26. Astronomical problems. One of the most important applications of Spherical Trigonometry is to Astronomy. In fact, Trigonometry was first developed by astronomers, and for centuries was studied only in connection with Astronomy. We shall take up the study of a few simple problems in Astronomy. 27. The celestial sphere. When there are no clouds to obstruct the view, the sky appears like a great hemispherical vault, with the observer at the center. The stars seem to glide upon the inner sur- face of this sphere from east to west,* their paths being parallel cir- cles whose planes are perpendicular to the polar axis of the earth, and having their centers in that axis produced. Each star | makes a complete revolution, called its diurnal (daily) motion, in 23 hr. 56 min., ordinary clock time. We cannot estimate the distance of the surface of this sphere from us, further than to perceive that it must be very far away indeed, because it lies beyond even the remotest terrestrial objects. To an observer the stars all seem to be at the same enormous distance from him, since his eyes can judge their directions only and not their distances. It is therefore natural, and it is extremely convenient from a mathematical point of view, to regard this imaginary sphere on which all the heavenly bodies seem to be projected, as having a radius of unlimited length. This * This apparent turning of the sky from east to west is in reality due to the rotation of the earth in the opposite direction, just as to a person on a swiftly moving train the objects outside seem to be speeding by, while the train appears to be at rest. The sky is really mo- tionless, while the earth is rotating from west to east. t By stars we shall mean fixed stars and nebula? whose relative positions vary so slightly that it takes centuries to make the change perceptible. 236 SPHERICAL TRIGONOMETRY sphere, called the celestial sphere, is conceived of as having such enormous proportions that the whole solar system (sun, earth, and planets) lies at its center, like a few particles of dust at the center of a great spherical balloon. The stars seem to retain the same relative positions with respect to each other, being in this respect like places on the earth's surface. As viewed from the earth, the sun, moon, planets, and comets are also projected on the celestial sphere, but they are changing their apparent positions with respect to the stars and with respect to each other. Thus, the sun appears to move east- ward with respect to the stars about one degree each day, while the moon moves about thirteen times as far. The following figure represents the celestial sphere, with the earth at the center showing as a mere dot. ( North celestial "i •pole) (North point AT ; — of 1VK horizon) ( South S point of horizon) 'p' (South celestial pole) Z' (Nadir) The zenith of an observer is the point on the celestial sphere directly overhead. A plumb line held by the observer and extended upwards will pierce the celestial sphere at his zenith {Z in figure). The nadir is the point on the celestial sphere which is diametric- ally opposite to the zenith (Z' in the figure). The horizon of an observer is the great circle on the celestial sphere having the observer's zenith for a pole ; hence every point on the horizon (SWNE in the figure) will be 90° from the zenith and from the nadir. A plane tangent* to a surface of still water * On account of the great distance, a plane passed tangent to the earth at the place of the observer will cut the celestial sphere in a great circle which (as far as we are concerned) coincides with the observer's horizon. APPLICATIONS OF SPHERICAL TRIGONOMETRY 237 at the place of the observer will cut the celestial sphere in his horizon. All points on the earth's surface have different zeniths and horizons. Every great circle passing through the zenith will be perpendicular to the horizon ; such circles are called vertical circles (as ZMHZ' and ZQSP'Z' in figure). The celestial equator or equinoctial is the great circle in which the plane of the earth's equator cuts the celestial sphere (EQWQ' in the figure). The poles of the celestial equator are the points (P and P' in the figure) where the earth's axis, if produced, would pierce the celestial sphere. The poles may also be defined as those two points on the sky where a star would have no diurnal (daily) motion. The Pole Star is near the north celestial pole, being about 1|° from it. Every point on the celestial equator is 90° from each of the celestial poles. All points on the earth's surface have the same celestial equator and poles. The geographical meridian of a place on the earth was defined as that great circle of the earth which passes through the place and the north and south poles. The celestial meridian of a point on the earth's surface is the great circle in which the plane, of the point's geo- graphical meridian cuts the celestial sphere (ZQSP'Z'Q'NP in the figure). It is evidently that vertical circle of an observer which passes through the north and south points of his horizon. All points on the surface of the earth which do not lie on the same north-and-south line have different celestial meridians. The hour circle of a heavenly body is that great circle of the celes- tial sphere which passes through the body * and through the north and south celestial poles. In the figure PMDP' is the hour circle of the star M. The hour circles of all the heavenly bodies are contin- ually changing with respect to any observer. The spherical triangle PZM, having the north pole, the zenith, and a heavenly body at its three vertices, is a very important triangle in Astronomy. It is called the astronomical triangle. 28. Spherical coordinates. When learning how to draw (or plot) the graph of a function, the student has been taught how to locate a point in a plane by measuring its distances from two fixed and mutu- ally perpendicular lines called the axes of coordinates, the two dis- tances being called the rectangular coordinates of the point. « By this is meant that the hour circle passes through that point on the celestial sphere where we see the heavenly body projected. 238 SPHERICAL TRIGONOMETRY If we now consider the surface to be spherical instead of plane, a similar system of locating points on it may be employed, two fixed and mutually perpendicular great circles being chosen as reference circles, and the angular distances of a point from these reference circles being used as the spherical coordinates of the point. Since the reference circles are perpendicular to each other, each one of them passes through the poles of the other. In his study of Geography the student has already employed such a system for locating points on the earth's surface, for the latitude and longitude of a point on the earth are really the spherical coordi- nates of the point, the two reference circles being the equator and the zero meridian (usually the meridian of Greenwich). Thus, in the figure on p. 231, we may consider the spherical coordinates of Boston to be the arcs ML (west longitude) and LB (north latitude) ; and of Cape Town the spherical coordinates would be the arcs M T (east longitude) and TC (south latitude). Similarly, we have systems of spherical coordinates for determining the position of a point on the celestial sphere, and (Zenith) e Z we shall now take up ^^ ' ^ := \~-^^ the study of the more /A l \ ,-'" ]\ nnportant of these. J J i / \ // \ 29. The horizon and ( North „ K/ I i / \ I \ pole) r /t:JI I /Z \ / \ meridian system. In i /b "\ j /f V M \ this case the two fixed £/ _. -"-^;rTi£^-----/4-"-' ! -- \ and mutll ally perpen- J^l ^:<lK a r<!L.jy-..T ,~"Jo dicular great circles of . \ / _L_^ — reference are the hori- Horizon w Ml H zon of tne observer ( Sunset) (SHWNE) and his meridian (SM 2 ZPN), and the spherical coordinates of a heavenly body are its altitude and azimuth. The altitude of a heavenly body is its angular distance above the ■ horizon measured on a vertical circle from 0° to 90°.* Thus the altitude of the sun M is the arc HM. The distance of a heavenly body from the zenith is called its zenith distance {ZM in the figure), and it is evidently the complement of its altitude. The altitude of the zenith is 90°. The altitude of the sun at sunrise or sunset is zero. The azimuth of a heavenly body is the angle between its vertical circle and the meridian of the observer. This angle is usually * At sea the altitude is usually measured by the sextant, while on laud a surveyor's transit is used. APPLICATIONS OF SPHERICAL TRIGONOMETRY 239 measured along the horizon from the south point westward to the foot of the body's vertical circle.* Thus the azimuth of the sun M ' is the angle SZH, which is measured by the arc SH. The azimuth of the sun at noon is zero and at midnight 180°. The azimuth of a star directly west of an observer is 90°, of one north 180°, and of one east 270°. Knowing the azimuth and altitude (spherical coordinates) of a heavenly, body, we can locate it on the celestial sphere as follows. Erom the south point of the horizon, as S (which may be considered the origin of coordinates, since it is an intersection of the reference circles), lay off the azimuth, as SH. Then on the vertical circle passing through H lay off the altitude, as HM. The body is then located at M. Ex. 1. In each of the following examples draw a figure of the celestial sphere and locate the body from the given spherical coordinates. Azimuth A Ititude Azimuth Altitui (a) 45° 45° (J) 0° 0° 0») 60° 30° (k) 180° 0° (o) 90° 60° (1) 0° 90° (d) 120° 75° (m) 90° 0° (e) 180° 55° (n) 270° 0° (f) 225° 0° (o) 360° 0° (g) 300° 60° (P) 330° 45° (a) 315° 15° (q) 75° 75° (i) 178° 82° W 90° 90° Since any two places on the earth have, in general, different merid- ians and different horizons, it is evident that this system of spher- ical coordinates is purely local. The sun rises at M x on the eastern horizon (altitude zero), mounts higher and higher in the sky, on a circle (M X M 2 M^) parallel to the celestial equator, until it reaches the observer's meridian M 2 (at noon, when its altitude is a maximum), then sinks downward to M z and sets on the western horizon. Similarly, for any other heavenly body, so that all are continually changing their altitudes and azimuths. To an observer having the zenith shown in the figure, a star in the northern sky near the north pole will not set at all, and to the same observer a star near the south pole will not rise at all. If its path for one day were traced on the celestial sphere, it would be a circle (as ABC) with its center in the polar axis and lying in a plane parallel to the plane of the equator. * That is, azimuth is measured from 0° to 360° clockwise. 240 SPHERICAL TRIGONOMETRY • -, s 30. The equator and meridian system. In this case the two fixed and mutually perpendicular great circles of reference are the celestial equator (EQDWQ 1 ) and the meridian of the observer (NPZQSP'Z'Q 1 ) ; and the spherical coordinates of a heavenly body are its declination and hour angle. The declination of a heavenly body is its angular distance north or south of the celestial equator measured on the hour circle of the body from 0° to 90°.* Thus, in the figure, the arc DM is a measure of the north declination of the star M. North declination is always considered positive and south declination nega- tive. Hence the decli- nation of the north pole is + 90°, while that of the south pole is - 90°. The declinations of the sun, moon, and planets are continually changing, but the dec- lination of a fixed star changes by an exceedingly small amount in the course of a year. The angular distance of a heavenly body from the north celestial pole, measured on the hour circle of the body, is called its north polar distance (PM in figure). The north polar distance of a star is evidently the complement of its declination. The hour angle of a heavenly body is the angle between the merid- ian of the observer and the hour circle of the star measured 'west- ward, from the meridian from 0° to 360°. Thus, in the figure, the hour angle of the star M is the angle QPD (measured by the arc QD). This angle is commonly used as a measure of time, hence the name hour angle. Thus the star M makes a complete circuit in 24 hours ; that is, the hour angle QPD continually increases at the uniform rate of 360° in 24 hours, or 15° an hour. For this reason the hour angle of a heavenly body is usually reckoned in hours from * The declinations of the sun, moon, planets, and some of the fixed stars, for any time of the year, are given in the Nautical Almanac or American Epkemeris, published by the United States government. APPLICATIONS OP SPHEEICAL TRIGONOMETRY 241 to 24, one hour being equal to 15°.* When the star is at M x (on the observer's meridian) its hour angle is zero. Then the hour angle increases until it becomes the angle M V PM (when the star is at M). When the star sets on the western horizon its hour angle becomes M^PM* Twelve hours after the star is at M 1 it will be at M 3 , when its hour angle will be 180° (= 12 hours). Continuing on its circuit, the star rises at M 4 and finally reaches M 1} when its hour angle has become 360° (= 24 hours), or 0° again. Knowing the hour angle and declination (spherical coordinates) of a heavenly body, we can locate it on the celestial sphere as follows. From the point, as Q, where the reference circles intersect, lay off the hour angle (or arc), as QD. Then on the hour circle passing through D lay off the declination, as DM. The body is then located at M. Ex. 1. In each of the following examples draw a figure of the celestial sphere and locate the body from the given spherical coordinates. Hour angle Declination Hour angle Declination (a) 45° N. 30° U) 00° S. 45° (°) 60° N. 60° W 0° 0° (e) 90° S. 45° (1) 180° 0° (d) 120° S. 30° (m) 90° N. 90° («) 180° N. 50° (n) 270° 0° (f) 5hr. N. 75° (o) 12 hr. S. 10° (g) 15 hr. -25° (P) 3hr. + 80° (h) 6hr. + 79° (q) 9hr. -45° « Ohr. -90° M 20 hr. + 60° 31. Practical applications. Among the practical applications of Astronomy the most important are : (a) To determine the position of an observer on the surface of the earth {i.e. his latitude and longitude). (b) To determine the meridian of a place on the surface of the earth. (c) To ascertain the exact tim.e of day at the place of the observer. (d) To determine the position of a heavenly body. The first of these, when applied to the determination of the place of a ship at sea, is the problem to which Astronomy mainly owes its economic importance. National astronomical observatories have been * On account of the yearly revolution of the earth ahout the sun, It takes the sun about 4 minutes longer to make the circuit than is required by any particular fixed star. Hence the solar day is about 4 minutes longer than the sidereal (star) day, but each is divided into 24 hours ; the first giving hours of ordinary clock time, while the second gives sidereal hours, which are used extensively in astronomical work. When speaking of the sun's hour angle it shall be understood that it is measured in hours of ordinary clock time, while the hour angle of a fixed star is measured in sidereal hours. In either case 1 hour = 15°. 242 SPHERICAL TRIGONOMETRY established, and yearly nautical almanacs are being published by the principal nations controlling the commerce of the world, in order to supply the mariner with the data necessary to determine his position accurately and promptly. 32. Relation between the observer's latitude and the altitude of the celestial pole. To an observer on the earth's equator (latitude zero) the pole star is on the horizon ; that is, the altitude of the star is zero. If the observer is traveling northward, the pole star will grad- ually rise ; that is, the latitude of the observer and the altitude of the star are both increasing. Finally, when the observer reaches the north pole of the earth his latitude and the altitude of the star have both increased to 90°. The place of the pole in the sky then depends in some way on the observer's latitude, and we shall now prove that the altitude of a celestial pole is equal to the latitude of the observer. Let be the place of observation, say some place in the northern hemisphere ; then the angle QCO (or arc QO) measures its north lati- tude. Produce the earth's axis CP until it pierces the celestial sphere at the celestial north pole. A line drawn from in the direction (as OP 2 ) of the celestial north pole will be parallel to CP 1} since the celestial north pole is at an unlimited distance from the earth (see § 27, p. 235). The angle NOP?, measures the altitude of the north pole. But CO is perpendicular to ON and CQ is perpendicular -to OP 2 (since it is perpendicular to the parallel line CP-i) ; hence the angles NOP 2 and QCO are equal, and we find that the altitude of the pole as observed at is equal to the latitude of 0. 33. To determine the latitude of a place on the surface of the earth. If we project that part of the celestial sphere which lies above the APPLICATIONS OF SPHERICAL TRIGONOMETRY 243 Horizon O ( Observer) horizon on the plane of the observer's celestial meridian, the horizon will be projected into a line (as NS), and the upper half of the celes- tial equator will also be projected into a line (as OQ). From the last section we know that the latitude of the observer equals the altitude of the elevated celestial pole (arc NP in figure), or, what amounts to the same thing, equals the angular distance between the zenith and the celestial equator (arc ZQ in figure). If then the elevated pole could be seen as a definitely marked point in the sky, the observer's latitude would be found by simply measuring the angular distance of that pole above the horizon. But there are no fixed stars visible at the exact points where the polar axis pierces the celestial sphere, the so-called polar star being about 1^° from the celestial north pole. Following are some methods for determining the lati- tude of a place on the surface of the earth. First method. To determine latitude by observations on circumsolar stars. The most obvious method is to observe with a suitable instru- ment the altitude of some star near the pole (so near the pole that it never .sets ; as, for instance, the star whose path in the sky is shown as the circle ABC in figure, p. 238) at the moment when it crosses the meridian above the pole, and again 12 hours later, when it is once more on the meridian but below the pole. In the first case its elevation will be the greatest possible ; in the second, the least possible. The mean of the two observed altitudes is evidently the latitude of the observer. Thus, in the figure on this page, if NA is the maximum altitude and NB the minimum altitude of the star, then NA + NB = NP = altitude of pole = latitude of place of observation. Ex. 1. The maximum altitude of a star near the pole star was observed to be 54° 16', and 12 hours later its minimum altitude was observed to be 40° 24'. What is the latitude of the place of observation ? Solution. 64° 16' + 40° 24' = 94° 40'. Therefore 94° 40' = 47° 20' = altitude of north pole = north latitude of place of observation. 244 SPHERICAL TRIGONOMETRY Horizon ( Observer ) Second method. To determine latitude from the meridian altitude of a celestial body whose declination is known. The altitude of a star M is measured when it is on z M(.star) the observer's meridian. If we subtract this meridian altitude (arc SM in figure) from 90°, we get the star's zenith dis- tance (ZM). In the Nautical Almanac we now look up the star's declination at the same instant ; this gives us the arc QM. Adding the declination of the star to its zenith distance, we get QM + MZ = QZ = NP = altitude of pole = latitude of place. Therefore, when the observer is on the northern hemisphere and the star is on the meridian south of zenith, North latitude = zenith distance + declination* If the star is on the meridian between the zenith and the pole (as at M" t), we will have North latitude = NP = ZQ = QM" — ZM" = declination — zenith distance. M(Star) P(South pole) If the observer is on the southern hemisphere and the star M is on his meridian between the zenith and south pole, we would have South latitude = SP' = SM— MP' = SM- (90°- QM) = altitude — co-declination, if we consider only the numerical value of the declination. In working out examples the student should depend on the figure rather than try to memorize formulas to cover all possible cases. Ex. 2. An observer in the northern hemisphere measured the altitude of a star at the instant it crossed his celestial meridian south of zenith, and found it to be 63° 40'. The declination of the star for the same instant was given by the Nautical Almanac as 21° 15' N. What was the latitude of the observer ? * If the star is south of the celestial equator (as at M' ), the same rale will hold, for then the declination is negative (south), and the algebraic sum of the zenith distance and decli- nation will still give the arc QZ. t Maximum altitude, if a circumpolar star. APPLICATIONS OF SPHEEICAL TRIGONOMETRY 245 63° 40', M (Star) Solution. Draw the semicircle NZSO. Lay off the arc SM - which locates the star at M. Since the dec- lination of the star is north, the celestial equator may be located by laying off the arc M Q = declination = 21° 15' towards the south. The line QO will then be the pro- jection of the celestial equator, and OP, drawn perpendicular to QO, will locate the north pole P. Zenith distance = ZM = 90° - SM (alt.) = 90° - 63° 40' = 26° 20'. .-. North latitude of observer = NP = ZQ = ZM (zen. dist.) + MQ (dec.) = 26° 20' + 21° 15' = 47° 35'. Third method. To determine latitude when the altitude, declination, and hour angle of a celestial body are known. Referring to the astro- nomical (spherical) triangle PZM, we see that side MZ = 90°- HM (alt.) = co-altitude, the altitude of the star being found by measure- ment. Also side PM = 90° - DM (dec.) = co-declination, the declination of the star being found from the Nautical Almanac. Angle ZPM = hour angle, which is given. This hour angle will be the local time when the observation is made on the sun. We then have two sides and the angle opposite one of them given in the spherical triangle PZM. Solving this for the side PZ, by Case III, (a), p. 224, we get Latitude of observer = NP = 90° — PZ. Ex. 3. The declination of a star is 69° 42' N. and its hour angle 60° 44'. "What is the north latitude of the place if the altitude of the star is observed to be 49° 40' ? Solution. Referring to the above figure, we have, in this example, side MZ = co-alt. = 90° - 49° 40' = 40° 20', side PM = co-dec. = 90° - 69° 42' = 20° 18', angle ZPM = hour angle = 60° 44'. Solving for the side PZ by Case III, (a), p. 224, we get side PZ =47° 9'=co-lat. .-. 90° - 47° 9' = 42° 51' = north latitude of place. The angle MZP is found to be 27° 53'; hence the azimuth of the star (angle SZH) is 180° - 27° 53' = 152° 7'. Horizon 246 SPHERICAL TRIGONOMETRY EXAMPLES 1. The following observations for altitude have been made on some north circumpolar star. What is the latitude of each place ? Maximum altitude Minimum altitude North latitude (a) New York 50° 46' 30° 40' Ans. 40° 43' (b) Boston 44° 22' 40° 20' 42° 21' (c) New Haven 58° 24' 24° 10' 41° 17' (d) Greenwich 64° 36' 38° 22' 51° 29' (e) San Francisco 55° 6' 20° 30' 37° 48' (f) Calcutta 24° 18' 20° 48' 22° 33' 2. In the following examples the altitude of some heavenly body has been measured at the instant when it crossed the observer's celestial meridian. What is the latitude of the observer in each case, the declination being found from the Nautical Almanac ? Hemisphere Meridian altitude (a) Northern (b) Northern (c) Northern (d) Northern (e) Northern (f) Northern (g) Southern (h) Southern (i) Southern (j) Southern 60° 75° 40' 43° 27' 38° 6' 50° 28° 46' 67° 45° 26' 72° 22° 18' Declination N. 20° N. 32° 13' S. 10° 52' S. 44° 26' N. 62° N. 73° 16' S. 59° S. 81° 48' S. 8° * N. 46° 25' Body is S. of zenith S. of zenith S. of zenith S. of zenith N. of zenith N. of zenith S. of zenith S. of zenith N. of zenith N. of zenith Ans. Latitude 50° N. 46° 33' N. 35°41'N. 7° 28' N. " 22° N. 12° 2' N. 36° S. 37° 14' S. 26° S. 21° 17' S. 3. In the following examples the altitude of" some heavenly body not on the observer's celestial meridian has been measured. The hour angle and declination are known for the same instant. Find the latitude of the observer in each case. Hemisphere Altitude Declination Hour angle Latitude (a) Northern 40° N. 10° 50° Ans. 27° 2' N. (b) Northern 15° S. 8° 65° 35°38'N. (c) Northern 52° N. 19° 2hr. 48° 16' N. (d) Northern 64° 42' N. 24° 20' 345° 3°34'N. or 46° 36' N. (e) Northern 0° S. 5° 5hr. 71° 22' N. (f) Northern 25° 0° 21 hr. 53°18'N. (g) Northern 0° N. 11° 14' 68° 64' No solution (h) Northern 9° 26' 0° 72° 22' 57°14'N. (i) Southern 38° S. 12° 52° 33° 56' S. or 4° 8' S. (j) Southern 19° N. 7° 3hr. 52° 56' S. (k) Southern 46° 18' S. 15° 23' 326° 49° 14' S. (1) Southern 0° N. 14° 38° 72° 26' S. (m) Southern 57° 36' 0° 2hr. 12° 50' S. APPLICATIONS OP SPHEEICAL TEIGONOMETRY 247 34. To determine the time of day. A very simple relation exists between the hour angla of the sun and the time of day at any place. The sun appears to move from east to west at the uniform rate of 15° per hour, and when the sun is on the meridian of a place it is apparent noon at that place. Comparing, Hour angle of sun 0° 15° 30° 45° 90° 180° 195° 210° 270° 300° 360° Time of day Noon 1 P.M. 2 P.M. 3 P.M. 6 P.M. Midnight 1 A.M. 2 A.M. 6 A.M. 8 A.M. Noon The hour angle of the sun M is the angle at P in the astronomical (spherical) triangle PZM. We may find this hour angle (time of (North p pole) *2 (North) JJ S( South) day) by solving the astronomical triangle for the angle at P, provided we know three other elements of the triangle. 248 SPHEEICAL TRIGONOMETRY DM = declination of sun, and is found from the Nautical Almanac. .'. Side PM = 90° — DM = co-declination of sun. HM = altitude of sun, and is found by measuring the angular dis- tance of the sun above the horizon with a sextant or transit. .". side MZ = 90° — HM = co-altitude of sun. NP = altitude of the celestial pole = latitude of the observer (p. 243). .'. Side PZ = 90°— NP = co-latitude of observer. Hence we have Rule for determining the time of day at a place whose latitude is known, when the declination and altitude of the sun at that time and place are known. First step. Take for the three sides of a spherical triangle the co-altitude of the sun, the co-declination of the sun, the co-latitude of the place. Second step. Solve this spherical triangle for the angle opposite the first-mentioned side. This will give the hour angle in degrees of the sun, if the observation is made in the afternoon. If the observation is made in the forenoon, the hour angle will be 360° — the angle found. Third step. When the observation is made in the afternoon the time of day will be hour angle IS When the observation is made in the forenoon the time of day will be ( *our angle _ 12 y M Ex. 1. In New York (lat. 40°43'N.) the sun's altitude is observed to be 30° 40'. Having given that the sun's decimation is 10° N. and that the observa- tion is made in the afternoon, what is the time of day? Solution. First step. Draw the triangle. Side a = co-alt. = 90° - 30° 40' = 59° 20'. Side 6 = co-dec. = 90° - 10° = 80°. Sun PoU Side c = co-lat. = 90° - 40° 43' = 49° 17'. Second step. As we have three sides given, the solution of this triangle comes under Case I, (a), p. 217. But as we only want the angle A (hour angle), some APPLICATIONS OF SPHEEICAL TKIGONOMETEY 249 labor may be saved by using one of the formulas (18), (19), (20), pp. 211, 212. Let us use (18), a= 59° 20' 6= 80° c= 49° 17' . . /sin s sin (s — a) sin \a = -t/ — -A ' , \ sin o sin c log sin £ a: = £ [log sin s + log sin (s - a) - {log sin 6 + log sin c}]. 2s = 188° 37' s = 94° 19'. « - a = 34° 59'. log sin s = 9. 9988 log sin 6=9. 9934 log sin (s - a) = 9.7584 log sin e= -9.8797 log numerator = 19.7572 log denominator = 19.8731 log denominator = 19.8731 9.8841 2 | 19.8841 log sin £ a = 9.9421 Ja=61°4'. a = 122° 8'. .-. A = 180° - a = 57° 52' = hour angle of sun. Third step. Time of day = — p.m. = 3 hr. 51 min. p.m. Ans. 15 EXAMPLES 1. In Milan (lat. 45°30'N.) the sun's altitude at an afternoon observation is 26° 30'. The sun's declination being 8° S., what is the time of day ? Ans. 2 hr. 33 min. p.m. 2. In New York (lat. 40° 43' N.) a forenoon observation on the sun gives 30° 40' as the altitude. What is the time of day, the sun's declination being 10° S.? Ans. 9 hr. 46 min. a.m. 3. A mariner observes the altitude of the sun to be 60°, its declination at the time of observation being 6° N. If the latitude of the vessel is 12° S., and the ob- servation is made in the morning, find the time of day. Ans. 10 hr. 24 min. a.m. 4. A navigator observes the altitude of the sun to be 35° 23', its declination being 10° 48' S. If the latitude of the ship is 26° 13' N., and the observation is made in the afternoon, find the time of day. Ans. 2 hr. 45 min. p.m. 5. At a certain place in latitude 40° N. the altitude of the sun was found to be 41° If its declination at the time of observation was 20° N., and the obser- vation was made in the morning, how long did it take the sun to reach the meridian ? Ans. 3 hr. 31 min. 6. In London (lat. 51° 31' N.) at an afternoon observation the sun's altitude is 15° 40'. Find the time of day, given that the sun's declination is 12° S. Ans. 2 hr. 59 min. p.m. 7. A government surveyor observes the sun's altitude to be 21°. If the latitude of his station is 27° N. and the declination of the sun 16° N. , what is the time of day if the observation was made in the afternoon 1 Ans. 4 hr. 57 min. p.m. 8. The captain of a steamship observes that the altitude of the sun is 26° 30'. If he is in latitude 45° 30' K and the declination of the sun is 18° N., what is the time of day if the observation was made in the afternoon ? Ans. 4 hr. 41 min. p.m. 250 SPHEEICAL TKIGONOMETKY 35. To find the time of sunrise or sunset. If the latitude of the place and the declination of the sun is known, we have a special case of the preceding problem ; for at sunrise or sunset the sun is on the hori- zon and its altitude is zero. Hence the co-altitude, which is one side of the astronomical triangle, will be 90°, and the triangle will be a quadrantal triangle (p. 204). The triangle may then be solved by the method of the last section or as a quadrantal triangle. EXAMPLES 1. At what hour will the sun set in Montreal (lat. 45° 30' N.), if its declina- tion at sunset is 18° N. ? Ans. 7 hr. 17 min. p.m. 2. At what hour will the sun rise in Panama (lat. 8° 57' N.), if its declination at sunrise is 23° 2' S.? Ans. 6 hr. 15 min. a.m. 3. About the first of April of each year the declination of the sun is 4° 30' N. I ind the time of sunrise on that date at tie following places : (a) New York (lat. 40° 43' N.). Ans. 5 hr. 45 min. a.m. (b) London (lat. 51° 31' N.). 5 hr. 37 min. a.m. (c) St. Petersburg (lat. 60° N.). 5 hr. 29 min. a.m. (d) New Orleans (lat. 29° 58' N.). 5 hr. 50 min. a.m. (e) Sydney (lat. 33° 52' S.). 6 hr. 12 min. a.m. 36. To determine the longitude of a place on the earth. From the definition of terrestrial longitude given on p. 231 it is evident that the meridians on the earth are projected into hour circles on the celestial sphere. Hence the same angle (or arc) which measures the angle between the celes- tial meridians, (hour circles) of the place of observation and of Greenwich may be taken as a measure of the longitude of the place. Thus, in the fig- ure, if PQP' is the me- ridian (hour circle) of Greenwich and PDP' the meridian (hour circle) of the place of observation, then the angle QPD (or arc QD) measures the west longitude of the place. If PMP' is the hour circle of the sun, it is evident that APPLICATIONS OP SPHERICAL TRIGONOMETRY 251 angle QPM = hour angle of sun for Greenwich = local time at Greenwich ; angle DPM = hour angle of sun for observer = local time at place of observation. Also, angle QPM — angle DPM = angle QPD = longitude of place. Hence the lo'ngitude of the place of observation equals the differ- ence * of local times between the standard meridian and the place in question. Or, in general, we have the following Rule for finding longitude : The observer's longitude is the amount by which noon at Greenwich is earlier or later than noon at the place of observation. If Greenwich has the earlier time, the longitude of the observer is east ; if it has the later time, then the longitude is west. We have already shown (p. 248) how the observer may find his own local time. It then remains to determine the Greenwich time without going there. The two methods which follow are those in general use. First method. Find Greenwich time by telegraph (wire or wireless'). By far the best method, whenever it is available, is to make a direct telegraphic comparison between the clock of the observer and that of some station the longitude of which is known. The difference between the two clocks will be the difference in longitude of the two places. Ex. 1. The navigator on a battleship has determined his local time to be 2 hr. 25 min. p.m. By wireless he finds the mean solar time at Greenwich to be i hr. 30 min. p.m. What is the longitude of the ship ? Solution. Greenwich having the later time, 4 hr. 30 min. 2 hr. 25 min. Subtracting, 2 hr. 5 min. = west longitude of the ship. Eeducing this to degrees and minutes of arc, 2 hr. 5 min. 15 Multiplying, 31° 15' = west longitude of ship. Second method. Find Greenwich time from a Greemvich chronom- eter. The chronometer is merely a very accurate watch. It has been set to Greenwich time at some place whose longitude is known, and thereafter keeps that time wherever carried. * This difference in time is not taken greater than 12 hours. If a difference in time be- tween the two places is calculated to he more than 12 hours, we subtract it from 24 hours and use the remainder instead as the difference. 252 SPHERICAL TRIGONOMETRY Ex. 2 An exploring party have calculated their local time to be 10 hr. a.m. The Greenwich chronometer which they cany gives the time as 8 hr. 30 min. a.m. What is their longitude ? Solution. Greenwich has here the earlier time. 10 hr. 8 hr. 30 min. Subtracting, 1 hr. 30 min. = 22° 30' = east longitude. EXAMPLES 1. In the following examples we have given the local time of the observer and the Greenwich time at the same instant. Find the longitude of the observer in each case. Observer's Corresponding Longitude local time Greenwich time of observer (a) Noon. 3 hr. 30 min. p.m. Ans. 52°30'W. (b) Noon. 7 hr. 20 min. a.m. 70° E. (c) Midnight. 10 hr. 15 min. p.m. 20° 15' E. (d) 4 hr. 10 min. p.m. Noon. 62° 30' E. (e) 8 hr. 25 min. a.m. Noon. 53°45'W. (f) 9 hr. 40 mill. p.m. Midnight. 36° W. (g) 2,hr. 15 min. p.m. 11 hr. 20 min. a.m. 43°45'E. (h) 10 hr. 26 min. a.m. 5 hr. 16 min. a.m. 77°30'E. (i) 1 hr. 30 min. p.m. 7 hr. 45 min. p.m. 93° 45' W. (j) Noon. Midnight. 180° W. orE. (k)0hr. p.m. 6hr. a.m. 180° E. orW. (1) 5 hr. 45 min. a.m. 7 hr. 30 min. p.m. 153° 45' E. (m) 10 hr. 55 min. p.m. 8 hr. 35 min. a.m. 145° W. 2. If the Greenwich time is 9 hr. 20 min. p.m., January 24, at the same instant that the time is 3 hr. 40 min. a.m., January 25, at the place of observation, what is the observer's longitude ? Ans. 96° E. 3. The local time is 4 hr. 40 min. a.m., March 4, and the corresponding Green- wich time is 8 hr. p.m., March 3. What is the longitude of the place ? Ans. 130° E. 4. In the following examples we have given the local time of the observer and the local time at the same instant of some other place whose longitude is known. Eind the longitude of the observer in each case. Observer's Corresponding time and Longitude local time longitude of the other place of observer (a) 2hr. p.m. 5 hr. p.m. at Havana (long. 82° 23' W.) Ans. 127°23'W. (b) 10 hr. a.m. 3 hr. p.m. at Yokohama (long. 139° 41' E.) 64°41'E. (c) 5 hr. 20 min. p.m. 11 hr. 30 min. p.m. at Glasgow (long. 4° 16' W.) 96°46'W. (d) 8hr.25min.A.M. 6hr.35min. a.m. at VeraOuz(long.96°9'W.) 68°39'W. (e) 9hr. 45min. p.m. Midnight at Batavia (long. 106° 52' E.) 73°7'E. (f) 7hr. 40 min. p.m. Noon at Gibraltar (long. 5° 21' W.) 109° 39' E. (g) 4 hr. 50 min. p.m. Noon at Auckland (long. 174° 50' E.) 112°40'W. APPLICATIONS OF SPHERICAL TRIGONOMETRY 253 6. What is the longitude of each place mentioned in the examples on p. 240, the Greenwich time for the same instant being given below ? Example, p. 249 (a) Ex. 3 (b) Ex; 4 (c) Ex. 5 (d) Ex. 7 (e) Ex. 8 Greenwich time 2 hr. 12 min. p.m. i hr. 52 min. p.m. 6 hr. 9 min. a.m. 10 hr. 33 min. p.m. 6 hr. 25 min. p.m. Longitude of place Ans. 57° W. long, (vessel) 31° WW. long, (vessel) 50° E. long, (observer) 84° W. long, (surveyor) 26° W. long, (ship) 37. The ecliptic and the equinoxes. The earth makes a complete circuit around the sun in one year. To us, however, it appeal's as if the sun moved and the earth stood still, the (apparent) yearly path of the sun among the stars being a great circle of the celestial sphere which we call the ecliptic. Evidently the plane of the earth's orbit Earth cuts the celestial sphere in the ecliptic. The plane of the equator and the plane of the ecliptic are inclined to each other at an angle of about 23^° (= e), called the obliquity of the ecliptic (angle LVQ in figure). The points where the ecliptic intersects the celestial equator are called the equinoxes. The point where the sun crosses the celestial equator when moving northward (in the spring, about March 21) is called the vernal equinox, and the point where it crosses the celestial equator when moving southward (in the fall, about September 21) is called the autumnal equinox. If we project the points V and A in our figure on the celestial sphere, the point V will be projected in the vernal equinox and the point A in the autumnal equinox. 38. The equator and hour circle of vernal equinox system.* The two fixed and mutually perpendicular great circles of reference are in • Sometimes called the equator system. 254 SPHERICAL TRIGONOMETRY (North celestial polej P this case the celestial equator (QVQ') and the hour circle of the ver- nal equinox (PVP') f also called the equinoctial colure ; and the spher- ical coordinates of a heavenly body are its declination and right ascension. The declination of a heavenly body has al- ready been defined on p. 240 as its angular distance north or south of the celestial equator measured on the hour circle of the body from 0° to 90°, positive if north and negative if south. In the figure DM is the north decli- nation of the star M. The right ascension of a heavenly body is the angle bet-ween the hour circle of the body and the hour circle of the vernal equinox measured eastward from the latter circle from 0° to 360°, or in hours from to 24. In the figure, the angle VPD (or the arc VD) is the right ascension of the star M. The right ascensions of the sun, moon, and planets are continually changing.* The angle LVQ (= e) is the obliquity of the ecliptic (= 23£°). Ex. 1. In each of the following examples draw a figure of the celestial sphere and locate the body from the given spherical coordinates. Bight ascension Declination Bight ascension Declination (a)0° 0° (j) 90° 0° (b) 180° 0° (k) 270° 0° (c) 90° N. 90° (1) 90° S. 90° (d) 46° N. 45° (m) 46° S.45° (e) 60° N. 60° (n) 90° S. 30° (f) 120° + 30° (o) 240° + 60° (g) 300° -60° (p) 330° -45° (h) 12 hr. + 45° (q) 6 hr. + 15° (i) 20 hr. 0° (r) 9hr. -75° * The right ascensions of the sun, moon, and planets may be found in the Nautical Almanac for any time of the year. APPLICATIONS OF SPHERICAL TRIGONOMETRY 255 Pole Ex. 2. The right ascension of a planet is 10 hr. 40 min. and its declination S. 6°. Find the angular distance from this planet to a fixed star whose right ascension is 3 hr. 20 min. and decimation N. 48°. Solution. Locate the planet and the star on the celestial sphere. Draw the spherical triangle whose vertices are at the north pole, the planet, and the fixed star. Then Angle A = difference of right ascensions = 10 hr. 40 min. - 3 hr. 20 min. = 7 hr. 20 min. = 110°. Side 5 = co-declination of star = 90° _ 48° = 42°. Side c = co-declination of planet = 90° - (- 6°) = 96°. To find side a. As we have two sides and the included angle given, the solution of this triangle comes under Case II, (a), p. 219. Since a only is required, the shortest method is that illustrated on p. 220, the solution depending on the solution of right spherical triangles. On solving, we get a= 107° 48'. Ans. 39. The system having for reference circles the ecliptic and the great circle KVK' passing through the pole of the ecliptic and the vernal (Sorth celestial pole) P Planet equinox.* The spherical coordinates of a heavenly body in this case are its latitude and longitude.^ The latitude of a heavenly body is its angular distance north or south of the ecliptic, measured on the great circle passing through * Sometimes called the ecliptic system. t Sometimes called celestial latitude and longitude in contradistinction to the latitude and longitude of places on the earth's surface (terrestrial latitude and longitude), which were defined on p. 231, and which have different meanings. 256 SPHEEICAL TRIGONOMETRY the body and the pole of the ecliptic. Thus, in the figure, the arc TM measures the north latitude of the star M. The longitude of a heavenly body is the angle between the great circle passing through the body and the pole of the ecliptic, and the great circle passing through the vernal equinox and the pole of the ecliptic, measured eastward from the latter circle from 0° to 360°. In the figure, the angle VKT (or the arc VT) is the longitude of the star M. The latitudes and longitudes of the sun, moon, and planets are continually changing. The angle LVQ (= e) is the obliquity of the ecliptic (= 23£° = arc KP). Since the ecliptic is the apparent yearly path of the sun, the celes- tial latitude of the sun is always zero. The declination of the sun, however, varies from N. 23£° (= arc QV) on the longest day of the year in the northern hemisphere (June 21), the sun being then the highest in the sky (at L), to S. 23£° (arc Q'L') on the shortest day of the year (December 22), the sun being then the lowest in the sky (at L'). The declination -of the sun is zero at the equinoxes (March 21 and September 21). Ex. 1. In each of the following examples draw a figure of the celestial sphere and locate the body from the given spherical coordinates. Celestial longitude Celestial latitude Celestial longitude Celestial latitude (a) 0° 0° (j) 90° 0° (b) 90° N. 90° (k) 180° 0° (c) 180° N. 45° (1) 0° S. 60° (d) 270° 0° (m) 60° N. 30° (e) 45° S. 30° (n) 120° N. 45° (f) 135° + 15° (o) 270° -75° (g) 315° + 60° (p) 30° -60° (h) 6hr. -45° (q) 9hr. 0° (i) 15 hr. + 45° (r) 18 hr. + 30° Ex. 2. Given the right ascension of a star 2 hr. 40 min. and its declination 24° 20' N., find its celestial latitude and longitude. Solution. Locate the star on the celestial sphere. Consider the spherical triangle KPM on the next page. Angle KPM = Z Q'PV + Z VPD = 90° + right ascension = 90° + 2 hr. 40 min. = 90° + 40° = 130°. Side PM = co-declension = 90°-24°20' = 65° 40'. and APPLICATIONS OP SPHEEICAL TRIGONOMETRY 257 Side KP = LQ = e = 23° 30'. To find side KM = co-latitude of the star, angle PKM = co-longitude of the star. (North pole) P As we have two sides and the included angle given, the solution of this tri- angle comes under Case II, (a), p. 219. Solving, we get Side KM = 81° 52' and Z PKM = 44° 52'. .-. 90° - KM = 90° - 81° 52' = 8° 8' = TM = latitude of star, and 90° - Z PKM = 90° - 44° 52' = 45° 8' = VT = longitude of star. EXAMPLES 1. Find the distance in degrees between the sun and the moon when their right ascensions are respectively 12 hr. 39 min., 6 hr. 56 min., and their declina- tions are 9° 23' S., 22° 50' N. Ans. 90°. 2. Find the distance between Regulus and Antares, the right ascensions being 10 hr. and 16 hr. 20 min.. and the polar distances 77° 19' and 116° 6'. Ans. 99° 56'. 3. Find the distance in degrees between the sun and the moon when their right ascensions are respectively 15 hr. 12 min., 4 hr. 45 min., and their decli- nations are 21° 30' S., 5° 30' N. Ans. 152° 23'. 4. The right ascension of Sirius is 6 hr. 39 min., and his declination is 16°31'S.; the right ascension of Aldebaran is 4hr. 27 min., and his declination is 16° 12' N. Find the angular distance between the stars. Ans. 46° 8'. 5. Given the right ascension of a star 10 hr. 50 min., and its declination 12° 30' N., find its latitude and longitude. Take e = 23° 30'. Ans. Latitude = 18° 24' N., longitude = 168° 53'. 6. If the moon's right ascension is 4 hr. 16 min. and its declination 6° 20' N., what is its latitude and longitude ? Ans. Latitude = 14° 43' N., longitude = 62° 58'. 258 SPHERICAL, TRIGONOMETRY 7. The sun's longitude was 69° 40'. What was its right ascension and decli- nation ? Take e = 23° 27'. Ans. Right ascension = 3 hr. 50 min., declination = 20° 5' N. Hint. The latitude of tlie sun is always zero, since it moves in the ecliptic. Hence in the triangle KPM (figure, p. 257), A"-l/= 90°, and it is a quadrantal triangle. This triangle may then be solved by the method explained on p. 204. 8. Given the sun's declination 16° 1' N. , find the sun's right ascension and longitude. Take e = io 3 27' Ans. Right ascension = 9 hr. 14 min., longitude = 136° 7'. 9. The sun's right ascension is 14 hr. 8 min. ; find its longitude and declina- tion. Take e = 23° 27'. Ans. Longitude = 214° 16', declination = 12° 56' S. 10. Find the length of the longest day of the year in latitude 42° 17' N. Ans. 15 hr. 6 min. Hint. This will he the time from sunrise to sunset when the sun is the highest in the sky, that is, when its declination is 23° 27' X. 11. Find the length of the shortest day in lat. 42° 17' N. Ans. 8 hr. 64 min. Hint. The sun will then he the lowest in the sky, that is, its declination will he 23° 27' S. 12. Find the length of the longest day in New Haven (lat. 41° 19' N.). Take e = 23° 27'. Ans. 15 hr. 13. Find the length of the shortest day in New Haven. Ans. 9 hr. 14. Find the length of the longest day in Stockholm (lat. 59° 21' N.). Take e = 23° 27'. Ans. 18 hr. 16 min. 15. Find the length of the shortest day in Stockholm. Ans. 5 hr. 48 min. 40. The astronomical triangle. TVe have seen that many of our most important astronomical problems depend on the solution of (North pole) t\ (North) jy S( South) the astronomical triangle PZM. In any such problem the first thing to do is to ascertain which parts of the astronomical triangle APPLICATIONS OF SPHERICAL TRIGONOMETRY 259 are given or can be obtained directly from the given data, and which are required. The different magnitudes which may enter into such problems are HM = altitude of the heavenly body, DM = declination of the heavenly body, angle ZPM = hour angle of the heavenly body, angle SZM = azimuth of the heavenly body, NP = altitude of the celestial pole = latitude of the observer. * As parts of the astronomical triangle PZM we then have side MZ = 90° - HM = co-altitude, side PM = 90° — DM = co-declination, side PZ = 90° - NP = co-latitude, angle ZPM = hour angle, angle PZM = 180° — azimuth (angle SZM)* The. student should be given practice in picking out the known and unknown parts in examples involving the astronomical tri- angle, and in indicating the case under which the solution of the triangle comes. For instance, let us take Ex. 15, p. 261. r Latitude = 51° 32' N .-. side PZ = 90°- 51° 32'= 38° 28'. Altitude = 35° 15'. .-. side ikfZ = 90°-35°15'=54°45'. Declination = 21° 27' N. .-. side MP = 90°- 21° 27'= 68° 33'. Required : Local time = hour angle = angle ZPM. Since we have three sides given to find an angle, the solution of the triangle comes under Case I, (a), p. 217. This gives angle ZPM = 59° 45'= 3 hr. 59 min. p.m. 41. Errors arising in the measurement of physical quantities. f Errors of some sort will enter into all data obtained by measurement. For instance, if the length of a line is measured by a steel tape, account must be taken of the expansion due to heat as well as the sagging of the tape under various tensions. Or, suppose the navigator of a ship * When the heavenly body is situated as in the figure. If the body is east of the ob- server's meridian, we would have angle PZM= azimuth — 180°. \ 1 Iu this connection the student is advised to read § 93 in Granville's Plane Trigonometry. Given parts 260 SPHERICAL TRIGONOMETRY at sea is measuring the altitude of the sun by means of a sextant. The observed altitude should be corrected for errors due to the following causes : 1. Dip. Owing to the observer's elevation above the sea level (on the deck or bridge of the ship), the observed altitude will be too great on account of the dip (or lowering) of the horizon. 2. Index error of sextant. As no instrument is perfect in con- struction, each one is subject to a certain constant error which is determined by experiment. S. Refraction of light. Celestial bodies appear higher than they really are because of the refraction of light by the earth's atmos- phere. This refraction will depend on the height of the celestial body above the horizon, and also on the state of the barometer and thermometer, since changes in the pressure and temperature of the air affect its density. Jf. Semidiameter of the sun. As the observer cannot be sure where the center of the sun is, the altitude of (say) the lower edge' of the sun is observed and to that is added the known semidiameter of the sun for that day found from the Nautical Almanac. 5. Parallax. The parallax of a celestial body is the angle sub- tended by the radius of the earth passing through the observer, as seen from the body. As viewed from the earth's surface, a celestial body appears lower than it would be if viewed from the center, and this may be shown to depend on the parallax of the body. We shall not enter into the detail connected with these correc- tions, as that had better be left to works on Eield Astronomy ; our purpose here is merely to call the attention of the student to the necessity of eliminating as far as possible the errors that arise when measuring physical quantities. For the sake of simplicity we have assumed that the necessary cor- rections have been applied to the data given in the examples found in this book. MISCELLANEOUS EXAMPLES 1. The continent of Asia has nearly the shape of an equilateral triangle. Assuming each side to be 4800 geographical miles and the radius of the earth to be 3440 geographical miles, find the area of Asia. Ans. About 13,333,000 sq. mi. 2. The distance between Paris (lat. 48° 50' N.) and Berlin (lat. 52° 30' N.) is 472 geographical miles, measured on the arc of a great circle. What time is it at Berlin when it is noon at Paris ? Ans. 44 rain, past noon. 3. The altitude of the north pole is 46°, and the azimuth of a star on the horizon is 135°. Find the polar distance of the star. Ans. 60°. APPLICATIONS OF SPHERICAL TRIGONOMETRY 261 4. What will be the altitude of the sun at 9 a.m. in Mexico City (lat. 19°26'N.), if its declination at that time is 8° 23' N. ? Ans. 45° 5'. 5. Find the altitude of the sun at 6 hr. a.m. at Munich (lat. 48° 9' N.) on the longest day of the year. Ans. Altitude = 17° 16'. 6. Find the time of day when the sun bears due east and due west on the longest day of the year at St. Petersburg (lat. 59° 56' N.). Ans. 6 hr. 58 min. a.m., 5 hr. 2 min. p.m. 7. What is the direction of a wall in lat. 52° 30' N. which casts no shadow at 6 a.m. on .the longest day of the year ? Ans. 75° 1 1', reckoned from the north point of the horizon. 8. Find the latitude of the place at which the sun rises exactly in the north- east on the longest day of the year. Ans. 65° 45' N. 9. Find the latitude of the place at which the sun sets at 10 hr. p.m. on the longest day. Ans. 63° 23' N. or S. 10. Given the latitude of the place of observation 52° 30' N. , the declination of a star 38°, its hour angle 28° 17'. Find the altitude of the star. Ans. Altitude = 65° 33'. 11. Given the latitude of the place of observation 51° 19' N., the polar dis- tance of a star 67° 59', its hour angle 15° 8'. Find the altitude and azimuth of the star. Ans. Altitude = 58° 23', azimuth = 27° 30'. 12. Given the declination of a star 7° 54' N. , its altitude 22° 45', its azimuth 50° 14'. Find the hour angle of the star and the latitude of the observer. Ans. Hour angle = 45° 41', latitude = 67° 59' N. 13. The latitude of a star is 51° N., and its longitude 315°. Find its declina- tion. Take e = 23° 27'. Ans. Declination = 32° 23' N. 14. Given the latitude of the observer 44° 50' N., the azimuth of a star 41° 2', its hour angle 20°. Find its declination. Ans. Declination = 20° 49' N. 15. Given the latitude of the place of observation 51° 32' N. , the altitude of the sun west of the meridian 35° 15', its declination 21° 27' N. Find the local time. Ans. 3 hr. 69 min. p.m. CHAPTER IV RECAPITULATION OF FORMULAS Spherical Trigonometry 42. Right spherical triangles, pp. 196-197. (1) cos c = cos a cos b, (2) sin a = sin c sin A, (3) sin b = sin sin B, (4) cos A = cos a sin B, (5) cos B = cos 6 sin A, (6) cos ^4 = tan b cot c, (7) cos B = tan a cot c, (8) sin b = tan a cot A, (9) sin a = tan b cot /J, (10) cos c = cot ,1 cot B. General directions for solving right spherical triangles by Napier's rules of circular parts are given on p. 200. Spherical isosceles and quadrantal triangles are discussed on p. 204. 43. Relations between the sides and angles of oblique spherical tri- angles, pp. 206-216. a = 180°-^, /J = 180°-j3, y = 180°-r;. d = diameter of inscribed circle. S = 180° — diameter of circumscribed circle. or Law of sines , p. 207. (ii; sin a. sini sin B sine sin^l sinC sin a sin b sine sin a: sin/3 siny Law of cosines for the sides, p. 209. (12) cos a = cos b cos « — sin b sin c cos a, 2fi2 RECAPITULATION OF FORMULAS Law of cosines for the angles, p. 209. (15) cos a = cos B cos y — sin 8 sin y cos a. Functions of ^ a, ^ B, £ y in terms of the sides, pp. 211-213 /iq\ - . /sin * sin (s — o) (18) smj«= A/ r-rA '- \ sin 6 sine 263 (19) (20) (27) {•28) (29) (30) cos tan i a . ^ J*^~( s - fe)sin(s -^7) \ sin b sin <r , a ^ r~sin^sin(g - a)~~ \sin (s — 6)sin(s — c) tan J<2 = -v sin (s — a) sin(s — q)sin(s — 6)sin(s — e) sins tan ^ a tanjjg tan J y = tan \d sin(s — b) tan | rf sin (s — c) tan Jd Functions of the half sides in terms of a, B, y, p. 214. (31) sin I sin a- sin (<r — a £a = "V ^-^ s > sin B sin y ). (32) (33) (40) (41) (42} (43) 1 sin j8 sin -y sin <r sin (a- — «) cos J tan * a = \iinT<r-i8)sin(«r-y) sin (a- — g)sin(a- — ft)sin(tr — y) sina- tan^a = tanji = tan j = sin (<r — a) tanjS sin (o- — B) tan^S sin (o- — y) tan JS 264 SPHEEICAL TRIGONOMETRY Napier's Analogies, p. 215. (44) tan i (a-i)=-^Jfc|l taili< .. v ' sin \ (a + /3) 2 (45) tan }(« + b) = - C ° S * ^ ~ g tan fr C . v ' v y cos i (a + /3) a (46) tan i(g - /?) =~ Sln t^ 7 ?? tan jy- v ' * v ^ y sin J (a + *) 2 ' (47) tanK« + ffl=- C0S tl a I?i tai1 ^- v ' a v ^ cos £ (a + b) 2 ' 44. General directions for the solution of oblique spherical triangles, pp. 216-227. Case 1. (a) Given ths three sides, p. J 17. (V) Given the three angles, p. 218. Case II. (a) Given two sides (uid their included angle, p. 219. (b) Given tiro angles and their included side, p. 222. Case III. (a) Given tiro sides and the angle opposite one of them p. 224. (b) Given tiro angles and the side opposite one of them p. 226. 45. Length of an arc of a circle in linear units, p. 228. {5Z) L ~ 180 N = number of degrees in angle. 46. Area of a spherical triangle, p. 229. (54) A!ea = l!r E=A+B + C -180°. (55) tan^ E = Vtan £ s tan ^ (s — a) tan ±(s — b) tan %(s — c) FOUR-PLACE TABLES OF LOGARITHMS COMPILED BY WILLIAM ANTHONY GEANVILLE, Ph.D., LLD. PBESIDENT OF PENNSYLVANIA COLLEGE GINN AND COMPANY BOSTON • NEW YORK • CHICAGO • LONDON ATLANTA ■ DALLAS • COLUMBUS ■ SAN FRANCISCO Entered at Statioxebs' Haix COPYRIGHT, 1908, BY WILLIAM ANTHONY GKANVILLE ALL RIGHTS RESERVED 715.12 gfct gtfcen«ii« >ctt« GINN AND COMPANY • PRO- PRIETORS • BOSTON ■ U.S.A. CONTENTS Pages Table I. Logarithms of Numbers 1-5 EULES FOR FINDING THE LOGARITHMS OF THE TRIGONO- METRIC Functions of Angles near 0° and 90° . . 6 Table II. Logarithms of the Trigonometric Func- tions, the Angle being expressed in Degrees and Minutes 7-16 Conversion Tables for Angles 17 Table III. Logarithms of the Trigonometric Func- tions, the Angle being expressed in Degrees and the Decimal Part of a Degree 19-37 Table of Natural Values of the Trigonometric Functions for Every Degree 38 Table I FOUR-PLACE LOGARITHMS OP NUMBERS This table gives the mantissas of the common logarithms (base 10) of the natural numbers (integers) from 1 to 2000, calculated to four places of decimals. A logarithm found from this table by interpolation may be in error by one unit in the last decimal place. TABLE I. LOGARITHMS OF NUMBERS No. 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 No. 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 1761 0004 0048 0090 0133 0175 0216 0257 0298 0338 0378 0418 0457 0496 0535 0573 0611 0648 0686 0722 0759 0795 0831 0867 0903 0938 0973 1007 1041 1075 1109 1143 1176 1209 1242 1274 1307 1339 1370 1402 1433 1464 1495 1526 1556 1587 1617 1647 1676 1706 1735 1764 0009 0052 0095 0137 0179 0220 0261 0302 0342 0382 0422 0461 0500 0538 0577 0615 0652 0689 0726 0763 0799 0835 0871 0906 0941 0976 1011 1045 1079 1113 1146 1179 1212 1245 1278 1310 1342 1374 1405 1436 1467 1498 1529 1559 1590 1620 1649 1679 1708 1738 1767 0013 0056 0099 0141 0183 0224 0265 0306 0346 0386 0426 0465 0504 0542 0580 0618 0656 0693 0730 0766 0803 0839 0874 0910 0945 0980 1014 1048 1082 1116 1149 1183 1216 1248 1281 1313 1345 1377 1408 1440 1471 1501 1532 1562 1593 1623 1652 1682 1711 1741 1770 0017 0060 0103 0145 0187 0228 0269 0310 0350 0390 0430 0469 0508 0546 0584 0622 0660 0697 0734 0770 0806 0842 0878 0913 0948 0983 1017 1052 1086 1119 1153 1186 1219 1252 1284 1316 1348 1380 1411 1443 1474 1504 1535 1565 1596 1626 1655 1685 1714 1744 1772 0022 0065 0107 0149 0191 0233 0273 0314 0354 0394 0434 0473 0512 0550 0588 0626 0663 0700 0737 0774 0026 0030 0069 0111 0154 0195 0237 0278 0318 0358 0398 0438 0477 0515 0554 0592 0630 0667 0704 0741 0777 0810 0813 0846 0881 0917 0952 0986 1021 1055 1089 1123 1156 1189 1222 1255 1287 1319 1351 1383 1414 1446 1477 1508 1538 1569 1599 1629 1658 1688 1717 1746 1775 0849 0885 0920 0955 0990 1024 1059 1093 1126 1159 1193 1225 1258 1290 1323 1355 1386 1418 1449 1480 1511 1541 1572 1602 1632 1661 1691 1720 1749 1778 0073 0116 0158 0199 0241 0282 0322 0362 0402 0441 0481 0519 0558 0596 0633 0671 0708 0745 0781 0817 0853 0888 0924 0959 0993 1028 1062 1096 1129 1163 1196 1229 1261 1294 1326 1358 1389 1421 1452 1483 1514 1544 1575 1605 1635 1664 1694 1723 1752 1781 0035 0077 0120 0162 0204 0245 0286 0326 0366 0406 0445 0484 0523 0561 0599 0637 0674 0711 0748 0785 0821 0856 0892 0927 0962 0997 1031 1065 1099 1133 1166 1199 1232 1265 1297 1329 1361 1392 1424 1455 1486 1517 1547 1578 1608 1638 1667 1697 1726 1755 1784 0039 0082 0124 0166 0208 0249 0290 0330 0370 0410 0449 0488 0527 0565 0603 0641 0678 0715 0752 0788 0824 0860 0896 0931 0966 1000 1035 1069 1103 1136 1169 1202 1235 1268 1300 1332 1364 1396 1427 1458 1489 1520 1550 1581 1611 1641 1670 1700 1729 1758 1787 Prop. Farts to a u s a ~v o& 1.0 16 2.0 2.6 3.0 3.5 4.0 4.5 4 1 0.4 2 0.8 3 1.2 4 1.6 5 2.0 6 2.4 7 2.8 H 3.2 9 3.6 3 1 0.3 2 0.6 3 OS) 4 1.2 5 1.5 6 1.8 7 2.1 8 2.4 9 2.7 2 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 TABLE I. LOGARITHMS OF XUMBEES Ho. 150 151 152 153 154 155 156 157 15S 159 160 161 162 163 164 165 166 167 16S 169 170 171 172 173 174 175 176 177 ITS 179 180 1S1 1S2 1S3 1S4 1S5 1S6 1S7 1SS 1S9 190 191 192 193 19+ 195 196 197 198 199 200 Ho. o ; i j 2 1761 1764 1790 ISIS 1847 1S75 1903 1931 1959 19S7 2014 2041 1767 1770 1772 1793 1796] 1S21 1S24 1S50J 1853 1S7S 1SS1 1906 1934 1962 1989 2017 2044 2068 2095 2122 214S 2175 2201 2227 2253 2279 1909' 1937 1965| 1992 j 2019 2047 2304 2071 209S 2125 2151 2177 2204 2230 2256 22S1 179S 1S27 1S55 1SS4I 1912 1940 1967 1995 2022 1801 1S30 1S5S 1SS6 1915 1942 1970 199S 2025 2049 2052 2307 2330 2355 j 23S0 | 2405 2430[ 2455 24S0J 2504 2529! 2074 2101 2127J 2154 2 ISO 2206 2232| 2258 22S4 1 2076 2079 2103) 2106 2130 2133 2156 21S3| 2209J 2235 2261| 22S7 2159 21S5 2212 22. 2263 22S9 23l0i 2312) ~23l5 iwo 235S 23S3 240S 2433 245S 24S2 2507 2531 i553j_: 25 2601 1 2625, 26481 2672 2695| 271SJ 2742 27651 2335 2360 23S5 2410 2435 2460 24S5 2509 2338 2363 23SS 2340 2365 2390 2413 2415 243S 2440 2463 2465 24S7 2512 2536 2490 2514 253S 255S, 2560 2579 25S2 2603 260i 2627 2629 2651 2674 2697 2721 2744 2767 27SS 2810 2S33 2856 287S 2900 2923 2945 2967 29S9 3010 2790 2S13 2S35 2S58 2S80 2903 2925 2947 2969 2991 3012 2562 2653 2676 2700 2723 2746 2769' 25S4 260S 2632 2655 2679 2702 2725 2749 2792 2794 2815 283S 2860 2SS3 2905 2927 2949 2971 2993| 2817 2840 2862 25S6 2610 2634 265S 26S1 2704 272S 2751 2774 2797 2S19 2S42 2865 2885 2SS7 2907 2909 2929: 2931 2951] 2953 2973 i 29! 299512997 30151 3017; 3019 8 i 9 1775 177S| 17S1 17S4 1787 1S04 1S07 1S10! 1S13; 1816 1S33 1S56 ! 1S3S 1S41I 1844 1S61 1S64 1S67! 1S70 1S72 1SS9S 1S92 1S95 1S9S 1917 1920 1923' 1926 19451 194S 1951 1953 1973, 2000| 202S 1976 2003 2030] 197S 2006 2033 2055; 2057, 2060 20S2 2109 2135 2162 21SS 2214 20S4 211l| 213S, 2164 1 219l| 2217 19S1 2009 2036 1901 1928 1956 19S4 2011 203S 1 2240; 2243 2266 2269 2292! 22941 20S 2114 2140 2167 21931 2219 2245 2271 2297i 2063 2066 2090 2117 2143' 21 70 1 2196! }>->■> 2IW 2274! 22991 231; 20 2343 236S| 2393 241S 2443 2467, 2492 2516 2541- 2345 2370 2395 2420 2445! 2470 2494 1 2519 2543 i 2348 2375' 239S! 2423' 244S 2472! 2325 2350 2375 2400 242. 2450 2475 I 2497! 2499 2521! 2524 2545 254S 2092 2119 2146 2172 2198 2225 2251 2276 2302 2327 2353 237S 2403 242S 245 2477 2502 2526 2550 2565: 2567 25 70 25 72 25S9 2613 2636 2660 1 26S3I 2707| 2730' 2732 2755 277S ; 2S01 2591 2615, 2639; 2662! 26S6I 2709 27761 2594 25961 2617 2620 2641 2643 2665| 2667 26SS 2690 2711 2714, 2735 2737! 275S 2760; 27Sl!27S3! 2574 259S 2622 2646 2669 2693 2716 2739 2762 27S5 2799' 2S04 2806 2522 2S44 2S67 2SS9 291l| 2934J 2956! 297S 2999 3021 2S24, 2S47 2869J 2S9ll 2914 2936| 295S' 2980! 3002! 2S26 2S49| 2S7l| 2S94 1 2916 293S 2S2S 2S51 2874 2S96 2918 2940 2960| 2962 29S2I29S4 3004 3006 3023 302S302S 2808 2831 2853 2876 2898 2920 2942 2964 2986 3008 3030 6 8 ! 9 Prop. Farts H 3 0.3 0.6 OS is 1.5 1.8 2J. 2.4 2.7 2 1 0.2 2 0.4 3 0.6 4 0.8 S 1.0 6 1.2 7 1.4 s 1.6 9 1.S TABLE I. LOGARITHMS OF NUMBERS No. 1 2 3 4 5 6 7 8 9 Prop. Farts 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 5S 59 60 61 62 63 64 65 66 67 68 69 70 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 bo •it •a d a M Difference 3222 3424 3617 3802 3979 4150 4314 4472 4624 3243 3444 3636 3820 3997 4166 4330 4487 4639 3263 3464 3655 3838 4014 4183 4346 4502 4654 3284 3483 3674 3856 4031 4200 4362 4518 4669 3304 3502 3692 3874 4048 4216 4378 4533 4683 3324 3522 3711 3892 4065 4232 4393 4548 4698 3345 3541 3729 3909 4082 4249 4409 4564 4713 3365 3560 3747 3927 4099 4265 4425 4579 4728 3385 3579 3766 3945 4116 4281 4440 4594 4742 3404 3598 3784 3962 4133 4298 4456 4609 4757 1 2 3 4 5 6 7 8 9 22 2.2 4.4 6.6 8.8 11.0 13.2 15.4 17.6 19.8 21 2.1 4.2 6.3 8.4 10.5 12.6 14.7 16.8 18.9 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 4914 5051 5185 5315 5441 5563 5682 5798 5911 4928 5065 5198 5328 5453 5575 5694 5809 5922 4942 5079 5211 5340 5465 5587 5705 5821 5933 4955 5092 5234 5353 5478 5599 5717 5832 5944 4969 5105 5237 5366 5490 5611 5729 5843 5955 4983 5119 5250 5378 5502 5623 5740 5855 5966 4997 5132 5263 5391 5514 5635 5752 5866 5977 5011 5145 5276 5403 5527 5647 5763 5877 5988 5024 5159 5289 5416 5539 5658 5775 5888 5999 5038 5172 5302 5428 5551 5670 5786 5900 6010 1 2 3 4 5 6 7 8 9 20 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 19 1.9 3.8 5.7 7.6 9Ji 11.4 13.3 15.2 17.1 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 1 2 3 4 5 6 7 8 9 18 1.8 3.6 5.4 7.2 9.0 10.8 12.6 14.4 16.2 17 1.7 3.4 5.1 6.8 8.5 10.2 11.9 13.6 15.3 6128 6232 6335 6435 6532 6628 6721 6812 6902 6138 6243 6345 6444 6542 6637 6730 6821 6911 6149 6253 6355 6454 6551 6646 6739 6830 6920 6160 6263 6365 6464 6561 6656 6749 6839 6928 6170 6274 6375 6474 6571 6665 6758 6848 6937 6180 6284 6385 6484 6580 6675 6767 6857 6946 6191 6294 6395 6493 6590 6684 6776 6866 6955 6201 6304 6405 6503 6599 6693 6785 6875 6964 6212 6314 6415 6513 6609 6702 6794 6884 6972 6222 6325 6425 6522 6618 6712 6803 6893 6981 1 2 3 4 5 6 7 8 9 16 1.6 3.2 4.8 6.4 8.0 9.6 11.2 12.8 14.4 15 1.5 3.0 4.5 6.0 7.5 9.0 10.5 12.0 133 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 7076 7160 7243 7324 7404 7482 7559 7634 7709 7084 7168 7251 7332 7412 7490 7566 7642 7716 7093 7177 7259 7340 7419 7497 7574 7649 7723 7101 7185 7267 7348 7427 7505 7582 7657 7731 7110 7193 7275 7356 7435 7513 7589 7664 7738 7118 7202 7284 7364 7443 7520 7597 7672 7745 7126 7210 7292 7372 7451 7528 7604 7679 7752 7135 7218 7300 7380 7459 7536 7612 7686 7760 7143 7226 7308 7388 7466 7543 7619 7694 7767 7152 7235 7316 7396 7474 7551 7627 7701 7774 1 2 3 4 5 6 7 8 9 14 1.4 2.8 4.2 5.6 7.0 8.4 9.8 11.2 12.6 13 1.3 2.6 3.9 5.2 6.5 7.8 9.1 10.4 11.7 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 7853 7924 7993 8062 8129 8195 8261 8325 8388 7860 7931 SOOO 8069 8136 8202 8267 833,1 8395 7868 7938 8007 8075 8142 8209 8274 8338 "8401 7875 7945 8014 8082 8149 8215 8280 8344 8407 7882 7952 8021 8089 8156 8222 8287 8351 8414 7889 7959 8028 8096 8162 8228 8293 8357 8420 7896 7966 8035 8102 8169 8235 8299 8363 8426 7903 7973 8041 8109 8176 8241 8306 8370 8432 7910 7980 8048 8116 8182 8248 8312 8376 8439 7917 7987 8055 8122 8189 8254 8319 8382 8445 1 2 3 4 5 6 7 8 9 12 1.2 2.4 3.6 4.8 6.0 7.2 8.4 9.6 10.8 11 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 No. 1 2 3 4 5 6 7 8 9 TABLE I. LOGARITHMS OF NUMBERS No. 1 2 3 4 5 6 7 8 9 Prop. Parts 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 Ex. dig. Difference 8513 8573 8633 8692 8751 8808 8865 S921 8976 8519 8579 8639 8698 8756 8814 8871 8927 8982 8525 8585 8645 8704 8762 8820 8876 8932 8987 8531 8591 8651 8710 8768 8825 8882 8938 8993 8537 8597 8657 8716 8774 8831 8887 8943 8998 8543 8603 8663 8722 8779 8837 8893 8949 9004 8549 8609 8669 8727 8785 8842 8899 8954 9009 8555 8615 8675 8733 8791 8848 8904 8960 9015 8561 8621 8681 8739 8797 8854 8910 8965 9020 8567 8627 8686 8745 8802 8859 8915 8971 9025 1 2 3 4 5 6 7 8 9 10 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 9 0.9 1.8 2.7' 3.6 4.5 5.4 6.3 7.2 8.1 1 2 3 i 5 6 7 8 9 1 2 3 4 5 6 7 8 9 8 0.8 1.6 2.4 3.2 4.0 4.8 5.6 6.4 7.2_ 6 0.6 1.2 1.8 2.4 3.0 3.6 4.2 4.8 5.4 7 0.7 1.4 2.1 2.8 3.5 4.2 4.9 5.6 6.3 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 9085 9138 9191 9243 9294 9345 9395 9445 9494 9090 9143 9196 9248 9299 9350 9400 9450 9499 9096 9149 9201 9253 9304 9355 9405 9455 9504 9101 9154 9206 9258 9309 9360 9410 9460 9509 9106 9159 9212 9263 9315 9365 9415 9465 9513 9112 9165 9217 9269 9320 9370 9420 9469 9518 9117 9170 9222 9274 9325 9375 9425 9474 9523 9122 9175 9227 9279 9330 9380 9430 9479 9528 9128 9180 9232 9284 9335 9385 9435 9484 9533 9133 9186 9238 9289 9340 9390 9440 9489 9538 5 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 9590 9638 9685 9731 9777 9823 9868 9912 9956 9595 9643 9689 9736 9782 9827 9872 9917 9961 9600 9647 9694 9741 9786 9832 9877 9921 9965 9605 9652 9699 9745 9791 9836 9881 9926 9969 9609 9657 9703 9750 9795 9841 9886 9930 9974 9614 9661 9708 9754 9800 9845 9890 9934 9978 0022 9619 9666 9713 9759 9805 9850 9894 9939 9983 9624 9671 9717 9763 9809 9854 9899 9943 9987 9628 9675 9722 9768 9814 9859 9903 9948 9991 9633 9680 9727 9773 9818 9863 9908 9952 9996 1 2 3 4 5 6 7 8 9 4 0.4 ' 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 0000 0004 0009 0013 0017 0026 0030 0035 0039 No. 1 2 3 4 5 6 7 8 9 RULES FOR FINDING THE LOGARITHMS OF THE TRIGONOMETRIC FUNCTIONS OF ANGLES NEAR 0° AND 90° The derivation of the following rules will be found on page 182, Granville's Plane Trigonometry. If the angle is given in degrees, minutes, and seconds, it should first be reduced to degrees and the decimal part of a degree. For this purpose use the conversion table on page 17. Rule I. To find the Logarithms of the Functions of an Angle near 0°.* log sin x° = 2.2419 + logx. log tan x° = 2.2419 + log x. log cot x° = 1.7581 - logx. log cos x° is found from the tables in the usual way. Rule II. To find the Logarithms of the Functions of an Angle near 90°. \ log cos x° = 2.2419 + log (90 — x). log cotx = 2.2419 + log (90 - x). log tan x° = 1.7581 - log (90 - x). log sin x° is found from the tables in the usual way. These rules ■will give results accurate to four decimal places for all angles between 0° and 1.1° and between 88.9° and 90°. * Example 1, page 182, Granville's Plane Trigonometry, illustrates the application of this rule. t Example 2, page 183, Granville's Plane Trigonometry, illustrates the application of this rule. Table II FOUR-PLACE LOGARITHMS OF TRIGONOMETRIC FUNCTIONS, THE ANGLE BEING EXPRESSED IN DEGREES AND MINUTES This table gives the common logarithms (base 10) of the sines, cosines, tangents, and cotangents of all angles from 0° to 5° and from 85° to 90° for each minute ; and from 5° to 85° at intervals of 10 minutes, all calculated to four places of decimals. In order to avoid the printing of negative characteristics, the number 10 has been added to every logarithm in the first, second, and fourth columns (those having log sin, log tan, and log cos at the top). Hence in writing down any logarithm taken from these three columns — 10 should be written after it. Logarithms taken from the third column (having log cot at the top) should be used as printed. A logarithm found from this table by interpolation may be in error by one unit in the last decimal place, except for angles between 0° and 18' or between 89° 42' and 90°, when the error may be larger. In the latter cases the table refers the student to the formulas on page 6 for more accurate results. TABLE II. LOGARITHMIC SINES 0° Angle log sin difi. 1' log tan com. difi.l' log cot log cos 0° 0' 0° 1' 10.0000 90° 00' 89° S9 7 6.4637 6.4637 3.5363 10.0000 0° 2' 6.7648 c 6.7648 S 3.2352 10.0000 89° 58' 0° 3' 6.9408 = 6.9408 " 3.0592 10.0000 89° 57' 0° 4' 7.0658 £ 7.0658 £ 2.9342 10.0000 89° 56' 0° 5' 7.1627 Ss°' 7.1627 '— — 2.8373 10.0000 89° 55' 0° 6' 7.2419 7.2419 2.7581 10.0000 89° 54' 0° 7' 7.3088 7.3088 s"o 2.6912 10.0000 89° 53' 0° 8' 7.3668 ts^ « 7.3668 si 3 2.6332 10.0000 89° 52' 0° 9' 0° 10' 0° 11' 7.4180 — ^ - a -- - u -_ i- « 3 o 7.4180 O P c a - = *« ° b -5' H ® <D ED 2.5820 10.0000 89° 51' 89° 50' 89° 49' 7.4637 7.4637 2.5363 10.0000 7.5051 7.5051 2.4949 10.0000 0° 12' 7.5429 7.5429 2.4571 10.0000 89° 48' 0° 13' 7.5777 7.5777 Sir* 2.4223 10.0000 89° 47' 0° 14' 7.6099 ■Elf 7.6099 ■sss 2.3901 10.0000 89° 46' 0° 15' 7.6398 7.6398 ogfl 2.3602 10.0000 89° 45' 0° 16' 7.6678 bo 7.6678 2.3322 10.0000 89° 44' 0° 17' 7.6942 7.6942 2.3058 10.0000 89° 43' 0° IS' 7.7190 7.7190 2.2810 10.0000 89° 42' 0° 19' 0°20' 0° 21' 7.7425 235 223 211 7.7425 235 223 212 2.2575 10.0000 89° 41' 89° 40' 89° 39' 7.7648 7.764S 2.2352 10.0000 7.7859 7.7S60 2.2140 10.0000 0° 22' 7.8061 202 7.8062 202 2.1938 10.0000 89° 38' 0° 23' 7.8255 194 7.8255 193 2.1745 10.0000 89° 37' 0° 24' 7.8439 184 7.8439 184 2.1561 10.0000 89° 36' 0° 25' 7.8617 178 7.8617 1(8 2.1383 10.0000 89° 35' 0° 26' 7.87S7 170 7.S787 170 2.1213 10.0000 89° 34' 0° 27' 7.8951 164 7.8951 164 2.1049 10.0000 89° 33' 0° 28' 7.9109 158 7.9109 158 2.0891 10.0000 89° 32' 0° 29' 0°30' 0° 31' 7.9261 152 147 143 7.9261 152 148 142 2.0739 10.0000 89° 31' 89° 30' 89° 2<y 7.940S 7.9409 2.0591 10.0000 7.9551 7.9551 2.0449 10.0000 0° 32' 7.9689 138 7.9689 138 2.0311 10.0000 S9° 28' 0° 33' 7.9822 133 7.9823 134 2.0177 10.0000 89° 27' 0° 34' 7.9952 130 7.9952 129 2.0048 10.0000 89° 26' 0° 35' 8.0078 126 8.0078 126 1.9922 10.0000 89° 25' 0° 36' 8.0200 122 8.0200 122 1.9800 10.0000 89° 24' 0° 37' 8.0319 119 8.0319 119 1.9681 10.0000 89° 23' 0° 3S' 8.0435 116 8.0435 116 1.9565 10.0000 10.0000 89° 22' 0° 39' 0°40' 0° 41' 8.0548 113 110 107 8.0548 113 110 107 1.9452 89° 21' 89° 20' 89° W 8.0658 8.0658 1.9342 10.0000 8.0765 8.0765 1.9235 10.0000 0° 42' 8.0870 105 8.0870 105 1.9130 10.0000 89° 18' 0° 43' 8.0972 102 8.0972 102 1.9028 10.0000 89° 17' 0° 44' 8.1072 100 8.1072 100 1.8928 10.0000 89° 16' 0° 45' 8.1169 97 8.1170 98 1.8830 10.0000 89° 15' 0° 46' 8.1265 96 8.1265 95 1.8735 10.0000 89° 14' 0° 47' 8.1358 93 8.1359 94 1.8641 10.0000 89° 13' 0° 48' 8.1450 92 8.1450 91 1.8550 10.0000 89° 12' 0° 49' 0°50' 0° 51' 8.1539 89 88 86 8.1540 90 87 86 1.8460 10.0000 89° 11' 89° 10' 89° 9' 8.1627 8.1627 1.8373 10.0000 8.1713 8.1713 1.8287 10.0000 0° 52' 8.1797 84 8.1798 85 1.8202 10.0000 89° 8' 0° 53' 8.1880 83 8.1880 82 1.8120 9.9999 89° 7' 0° 54' 8.1961 81 8.1962 82 1.8038 9.9999 89° 6' 0° 55' 8.2041 80 8.2041 79 1.7959 9.9999 89° 5' 0° 56' 8.2119 78 8.2120 79 1.7880 9.9999 89° 4' 0° 57' 8.2196 77 8.2196 76 1.7804 9.9999 89° 3' 0° 58' 8.2271 75 8.2272 76 1.7728 9.9999 89° 2' 0° 59' 0° 60' 8.2346 75 73 8.2346 74 73 1.7654 9.9999 89° 1' 89° C 8.2419 8.2419 1.7581 9.9999 log cos diff.l' log cot com. difi.l' log tan log sin Angle 89° COSINES, TANGENTS, AND COTANGENTS 1° Angle log sin diS.l' log tan com. difi.l' log cot log cos 1° 0' 1° 1' 8.2419 71 8.2419 72 1.7581 9.9999 88° 60' 88° 59' 8.2490 8.2491 1.7509 9.9999 1° 2' 8.2561 8.2562 71 1.7438 9.9999 88° 58' 1° 3' 8.2630 69 8.2631 69 1.7369 9.9999 88° 57' 1° 4' 8.2699 69 8.2700 69 1.7300 9.9999 88° 56' 1° 5' 8.2766 67 8.2767 67 1.7233 9.9999 88° 55' 1° 6' 8.2832 66 8.2833 66 1.7167 9.9999 88° 54' 1° 7' 8.2898 66 8.2899 66 1.7101 9.9999 88° 53' 1° 8' 8.2962 64 8.2963 64 1.7037 9.9999 88° 52' 1° 9' 1° 10' 1° 11' 8.3025 63 63 62 8.3026 63 63 61 1.6974 9.9999 S8° 51' 88° 50' 88° 49' 8.3088 8.3089 1.6911 9.9999 8.3150 8.3150 1.6850 9.9999 1° 12' 8.3210 60 8.3211 61 1.6789 9.9999 88° 48' 1° 13' 8.3270 60 8.3271 60 1.6729 9.9999 88° 47' 1° 14' 8.3329 59 8.3330 59 1.6670 9.9999 88° 46' 1° 15' 8.3388 59 8.3389 59 1.6611 9.9999 88° 45' 1° 16' 8.3445 57 8.3446 57 1.6554 9.9999 88° 44' 1° 17' 8.3502 57 8.3503 56 1.6497 9.9999 88° 43' 1° 18' 8.355S 56 8.3559 56 1.6441 9.9999 88° 42' 1° 19' 1°20' 1° 21' 8.3613 55 55 54 8.3614 55 55 54 1.6386 9.9999 88° 41' 88° 40' 88° 39' 8.3668 8.3669 1.6331 9.9999 8.3722 8.3723 1.6277 9.9999 1° 22' 8.3775 53 8.3776 1.6224 9.9999 88° 38' 1° 23' 8.3828 53 8.3829 53 1.6171 9.9999 88° 37' 1° 24' 8.3880 52 8.3881 52 1.6119 9.9999 88° 36' 1° 25' 8.3931 51 8.3932 51 1.6068 9.9999 88° 35' 1° 26' 8.3982 51 8.3983 51 1.6017 9.9999 88° 34' 1° 27' 8.4032 50 8.4033 50 1.5967 9.9999 88° 33' 1° 28' 8.4082 50 8.4083 50 1.5917 9.9999 88° 32' 1° 29' 1°30' 1° 31' 8.4131 49 49 48 8.4132 49 49 48 1.5868 9.9999 88° 31' 88° 30' 88° 29' 8.4179 8.4227 8.4181 1.5819 9.9999 8.4229 1.5771 9.9998 1° 32' 8.4275 48 8.4276 47 1.5724 9.9998 88° 28' 1° 33' 8.4322 47 8.4323 47 1.5677 9.9998 88° 27' 1° 34' 8.4368 46 8.4370 47 1.5630 9.9998 88° 26' 1° 35' 8.4414 46 8.4416 46 1.5584 9.9998 88° 25' 1° 36' 8.4459 45 8.4461 45 1.5539 9.9998 88° 24' 1° 37' S.4504 45 8.4506 45 1.5494 9.9998 88° 23' 1° 38' 8.4549 45 8.4551 45 1.5449 9.9998 88° 2'2' 1° 39' 1°40' 1° 41' 8.4593 44 44 43 8.4595 44 43 44 1.5405 9.9998 88° 21' 88° 20' 88° 19' 8.4637 8.4638 1.5362 9.9998 8.4680 8.4682 1.5318 9.9998 1° 42' 8.4723 43 8.4725 43 1.5275 9.9998 88° 18' 1° 43' 8.4765 42 8.4767 42 1.5233 9.9998 88° 17' 1° 44' 8.4807 42 8.4809 42 1.5191 9.9998 88° 16' 1° 45' 8.4848 41 8.4851 42 1.5149 9.9998 88° 15' 1° 46' 8.4890 42 8.4892 41 1.5108 9.9998 88° 14' 1° 47' 8.4930 40 8.4933 41 1.5067 9.9998 88° 13' 1° 48' 8.4971 41 8.4973 40 1.5027 9.9998 88° 12' 1° 49' 1°50' 1° 51' 8.5011 40 39 40 8.5013 40 40 39 1.4987 9.9998 88° 11' 88° 10' 88° 9' 8.5050 8.5053 1.4947 9.9998 8.5090 8.5092 1.4908 9.9998 1° 52' 8.5129 39 8.5131 39 1.4869 9.9998 88° 8' 1° 53' 8.5167 38 8.5170 39 1.4830 9.9998 88° 7' 1° 54' 8.5206 39 8.5208 38 1.4792 9.9998 88° 6' 1° 55' 8.5243 37 8.5246 38 1.4754 9.9998 88° 5' 1° 56' 8.5281 38 8.5283 37 1.4717 9.9998 88° 4' 1° 57' 8.5318 37 8.5321 38 1.4679 9.9997 88° 3' 1° 58' 8.5355 37 8.5358 37 1.4642 9.9997 88° 2' 1° 59' 1°60' 8.5392 37 ' 36 8.5394 36 37 1.4606 9.9997 88° 1' 88° 0' 8.5428 8.5431 1.4569 9.9997 log cos difi.l' log cot com. difi.l' log tan log sin Angle 88° 10 TABLE II. LOGARITHMIC SINES 2° Angle log sin difi.l' log tan dig}/ log cot log cos 0' 1' 2' 3' 4' 5' 6' 7' 8' •r 10' ir 2° 12' 2° 13' 2° 14' 2° IS' 2° 16' 2° 17' 2° 18' 2° Vf 2° 20' 2° 21' 2° 22' 2° 23' 2° 24' 2° 25' 2° 26' 2° 27' 2° 28' 2° 29' 2° 30' 2° 31' 2° 32' 2° 33' 2° 34' 2° 35' 2° 36' 2° 37' 2° 38' 2° 39' 2° 40' 2° 41' 2° 42' 2° 43' 2° 44' 2° 45' 2° 46' 2° 47' 2° 48' 2° 49' 2° 50' 2° 51' 2° 52' 2° 53' 2° 54' 2° 55' 2° 56' 2° 57' 2° 58' 2° 59' 2° 60' 8.5428 8.5464 8.5500 8.5535 8.5571 8.5605 8.5640 8.5674 8.5708 8.5742 8.5776 8.5809 8.5842 8.5875 8.5907 8.5939 8.5972 8.6003 8.6035 8.6066 8.6097 8.6128 8.6159 8.6189 8.6220 8.6250 8.6279 8.6309 8.6339 8.6368 8.6397 8.6426 8.6454 8.6483 8.6511 8.6539 8.6567 8.6595 8.6622 8.6650 8.6677 8.6704 8:6731 8.6758 8.6784 8.6810 8.6837 8.6863 8.6889 8.6914 8.6940 8.6965 8.6991 8.7016 8.7041 8.7066 8.7090 8.7115 8.7140 8.7164 8.7188 36 .16 35 36 34 35 34 34 34 34 33 33 33 33 32 33 31 32 31 31 31 31 30 31 30 29 30 30 29 29 29 28 29 28 27 28 27 27 27 27 27 26 26 25 26 25 26 25 25 25 24 25 25 24 24 8.5431 8.5467 8.5503 8.5538 8.5573 8.5608 8.5643 8.5677 8.5711 8.5745 8.5779 8.5812 8.5845 8.5878 8.5911 8.5943 8.5975 8.6007 8.6038 8.6070 8.6101 8.6132 8.6163 8.6193 8.6223 8.6254 8.6283 8.6313 8.6343 8.6372 8.6401 8.6430 8.6459 8.6487 8.6515 8.6544 8.6571 8.6599 8.6627 8.6654 8.6682 8.6709 8.6736 8.6762 8.6789 8.6815 8.6842 8.6868 8.6894 8.6920 8.6945 8.6971 8.6996 8.7021 8.7046 8.7071 8.7096 8.7121 8.7145 8.7170 8.7194 36 36 35 35 35 35 34 34 34 34 33 33 33 33 32 32 32 31 32 31 31 31 30 30 31 29 30 30 29 29 29 28 29 27 27 28 27 27 26 27 26 27 26 36 26 25 26 25 25 25 25 25 25 24 25 24 1.4569 1.4533 1.4497 1.4462 1.4427 1.4392 1.4357 1.4323 1.4289 1.4255 1.4221 1.4188 1.4155 1.4122 1.4089 1.4057 1.4025 1.3993 1.3962 1.3930 1.3899 1.3868 1.3837 1.3807 1.3777 1.3746 1.3717 1.3687 1.3657 1.3628 1.3599 1.3570 1.3541 1.3513 1.3485 1.3456 1.3429 1.3401 1.3373 1.3346 1.3318 1.3291 1.3264 1.3238 1.3211 1.3185 1.3158 1.3132 1.3106 1.3080 1.3055 1.3029 1.3004 1.2979 1.2954 1.2929 1.2904 1.2879 1.2855 1.2830 1.2806 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9997 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 log cos difi. 1' log cot fljg™ f' log tan log sin 8T 87° 60' 87° 59' 87° 58' 87° 57' 87° 56' 87° 55' 87° 54' 87° 53' 87° 52' 87° 51' 87° 50' 87° 49' 87° 48' 87° 47' 87° 46' 87° 45' 87° 44' 87° 43' 87° 42' 87° 41' 87° 40' 87° Z<¥ 87° 38' 87° 37' 87° 36' 87° 35' 87° 34' 87° 33' 87° 32' 87° 31' 87° 30' 87° 29' 87° 28' 87° 27' 87° 26' 87° 25' 87° 24' 87° 23' 87° 22' 87° 21' 87° 20' 87° 19' 87° 18' 87° 17' 87° 16' 87° 15' 87° 14' 87° 13' 87° 12' 87° 11' 87° 10' 87° 9' 87° 87° 87° 87° 87° 87° 87° 87° 87° Angle COSINES, TANGENTS, AND COTANGENTS 11 a o Angle log sin difi.l' log tan com. difi.l' log cot log cos 3° 0' 3° 1' 8.7188 24 8.7194 24 1.2806 9.9994 86° 60' 86° 59' 8.7212 8.7218 1.2782 9.9994 3° 2' 8.7236 24 8.7242 24 1.2758 9.9994 86° 58' 3° 3' 8.7260 24 8.7266 24 1.2734 9.9994 86° 57' 3° 4' 8.7283 23 8.7290 24 1.2710 9.9994. 86° 56' 3° 5' 8.7307 24 8.7313 23 1.2687 9.9994 86° 55' 3° 6' 8.7330 23 8.7337 24 1.2663 9.9994 86° 54' 3° 7' 8.7354 24 8.7360 23 1.2640 9.9994 86° 53' 3° 8' 8.7377 23 8.7383 23 1.2617 9.9994 86° 52' 3° 9' 3° 10' 3° 11' 8.7400 23 23 22 8.7406 23 23 23 1.2594 9.9993 86° 51' 86° 50' 86° 49' 8.7423 8.7429 1.2571 9.9993 8.7445 8.7452 1.2548 9.9993 3° 12' 8.7468 23 8.7475 23 1.2525 9.9993 86° 48' 3° 13' 8.7491 23 8.7497 22 1.2503 9.9993 86° 47' 3° 14' 8.7513 22 8.7520 23 1.2480 9.9993 86° 46' 3° 15' 8.7535 22 8.7542 22 1.2458 9.9993 86° 45' 3° 16' 8.7557 22 8.7565 23 1.2435 9.9993 86° 44' 3° 17' S.7580 23 8.7587 22 1.2413 9.9993 86° 43' 3° 18' 8.7602 22 8.7609 22 1.2391 9.9993 86° 42' 3° 19' 3° 20' 3° 21' 8.7623 21 22 22 8.7631 22 21 22 1.2369 9.9993 86° 41' 86° 40' 86° 39' 8.7645 8.7652 1.2348 9.9993 8.7667 8.7674 1.2326 9.9993 3° 22' 8.7688 2L 8.7696 22 1.2304 9.9993 86° 38' 3° 23' 8.7710 22 8.7717 21 1.2283 9.9992 86° 37' 3° 24' 8.7731 21 8.7739 22 1.2261 9.9992 86° 36' 3° 25' 8.7752 21 8.7760 21 1.2240 9.9992 86° 35' 3° 26' 8.7773 21 8.7781 21 1.2219 9.9992 86° 34' 3° 27' 8.7794 21 8.7802 21 1.2198 9.9992 86° 33' 3° 28' 8.7815 21 8.7823 21 1.2177 9.9992 86° 32' 3° 29' 3° 30' 3° 31' 8.7836 21 21 20 8.7844 21 21 21 1.2156 9.9992 86° 31' 86° 30' 86° 29' 8.7857 8.7865 1.2135 9.9992 8.7877 8.7886 1.2114 9.9992 3° 32' 8.7898 21 8.7906 20 1.2094 9.9992 86° 28' 3° 33' 8.7918 20 8.7927 21 1.2073 9.9992 86° 27' 3° 34' 8.7939 21 8.7947 20 1.2053 9.9992 86° 26' 3° 35' 8.7959 20 8.7967 20 1.2033 9.9992 86° 25' 3° 36' 8.7979 20 8.7988 21 1.2012 9.9991 86° 24' 3° 37' 8.7999 20 8.8008 20 1.1992 9.9991 86° 23' 3° 38' 8.8019 20 8.8028 20 1.1972 9.9991 86° 22' 3° 39' 3° 40' 3° 41' 8.8039 20 20 19 S.S048 8.S067 20 20 20 1.1952 9.9991 86° 21' 86° 20' 86° 19' 8.8059 1.1933 9.9991 8.8078 8.80S7 1.1913 9.9991 3° 42' 8.8098 20 8.8107 20 1.1893 9.9991 86° 18' 3° 43' 8.8117 19 8.S126 19 1.1874 9.9991 86° 17' 3° 44' 8.8137 20 8.8146 20 1.1854 9.9991 86° 16' 3° 45' 8.8156 19 8.8165 19 1.1835 9.9991 86° 15' 3° 46' 8.8175 19 8.8185 20 1.1815 9.9991 86° 14' 3° 47' 8.8194 19. 8.8204 19 1.1796 9.9991 86° 13' 3° 48' 8.8213 19 8.8223 19 1.1777 9.9990 86° 12' 3° 49' 3° 50' 3° 51' 8.8232 19 19 19 8.8242 19 19 19 1.1758 9.9990 86° 11' 86° 10' 86° 9' 8.8251 8.8261 1.1739 9.9990 8.8270 8.8280 1.1720 9.9990 3° 52' 8.8289 19 8.8299 19 1.1701 9.9990 86° 8' 3° 53' 8.8307 18 8.8317 18 1.1683 9.9990 86° 7' 3° 54' 8.8326 19 8.8336 19 1.1664 9.9990 86° 6' 3° 55' 8.8345 19 8.8355 19 1.1645 9.9990 86° 5' 3° 56' 8.8363 18 8.8373 18 1.1627 9.9990 86° 4' 3° 57' 8.8381 18 8.8392 19 1.1608 9.9990 86° 3' 3° 58' 8.8400 19 8.8410 18 1.1590 9.9990 86° 2' 3° 59' 3° 60' 8.8418 18 18 8.8428 18 18 1.1572 9.9990 86° 1' 86° 0' 8.8436 8.8446 1.1554 9.9989 log cos difi.l' log cot com. diff.1' log tan log sin Angle 86° 12 TABLE II. LOGARITHMIC SINES Angle 0' 1' 2' 3' 4' 5' 6' V 8' 9' 10' 11' 12' 4° 13' 4° 14' 4° 15' 4° 16' 4° 17' 4° 18' 4° 19' 4° 20' 4° 21' 4° '22' 4° 23' 4° 24' 4° 25' 4° 26' 4° 27' 4° 28' 4° 29' 4° 30' 4° 31' 4° 32' 4° 33' 4° 34' 4° 35' 4° 36' 4° 37' 4° 38' 4° 39' 4° 40' 4° 41' 4° 42' 4° 43' 4° 44' 4° 45' 4° 46' 4° 47' 4° 48' 4° 49' 4° 50' 4° 51' 4° 52' 4° 53' 4° 54' 4° 55' 4° 56' 4° 57' 4° 58' 4° 59' 4° 60' log sin (lift. 1 8.8436 8.8454 8.8472 8.8490 8.8508 8.8525 8.8543 8.8560 8.8578 8.8595 8.8613 8.8630 8.8647 8.8665 8.8682 8.8699 8.8716 8.8733 8.8749 8.8766 8.8783 8.8799 8.8816 8.8833 8.8849 8.8865 8.8882 8.8898 8.8914 8.8930 8.8962 8.8978 8.8994 8.9010 8.9026 8.9042 8.9057 8.9073 8.9089 8.9104 8.9119 8.9135 8.9150 8.9166 8.9181 8.9196 8.9211 8.9226 8.9241 8.9256 8.9271 8.9286 8.9301 8.9315 8.9330 8.9345 8.9359 8.9374 8.9388 8.9403 ]8 18 18 18 17 18 17 18 17 18 17 18 17 17 17 17 10 17 17 16 17 17 16 16 17 IG 10 10 16 16 16 10 16 10 10 15 10 10 15 15 16 15 10 15 15 15 15 15 15 15 15 15 14 15 15 14 15 15 15 log tan 8.8446 8.8465 8.8483 8.8501 8.8518 8.8536 8.8554 S.8572 8.8589 8.8607 8.8624 8.8642 8.8659 8.8676 8.8694 8.8711 8.8728 8.8745 8.8762 8.8778 8.8795 8.8812 8.8829 8.8845 8.8862 8.8878 8.8895 8.8911 8.8927 8.8944 8.8960 8.8976 8.8992 8.9008 8.9024 8.9040 8.9056 8.9071 8.9087 8.9103 8.9118 8.9134 8.9150 8.9165 8.9180 8.9196 8.9211 8.9226 8.9241 8.9256 8.9272 8.9287 8.9302 8.9316 8.9331 8.9346 8.9361 8.9376 8.9390 8.9405 8.9420 com. cliff. 1' 19 18 18 17 18 18 18 17 18 17 18 17 17 18 17 17 17 17 16 17 17 10 17 16 17 16 16 17 16 16 16 16 16 16 16 15 10 16 15 16 16 15 15 16 15 15 15 15 16 15 15 14 15 15 15 15 14 15 15 log cos diff.l' log cot d "jft™i 85° log cot 1.1554 1.1535 1.1517 1.1499 1.1482 1.1464 1.1446 1.1428 1.1411 1.1393 1.1376 1.1358 1.1341 1.1324 1.1306 1.1289 1.1272 1.1255 1.1238 1.1222 1.1205 1.1188 1.1171 1.1155 1.1138 1.1122 1.1105 1.1089 1.1073 1.1056 1.1040 1.1024 1.1008 1.0992 1.0976 1.0960 1.0944 1.0929 1.0913 1.0897 1.0882 1.0866 1.0850 1.0835 1.0820 1.0804 1.0789 1.0774 1.0759 1.0744 1.0728 1.0713 1.0698 1.0684 1.0669 1.0654 1.0639 1.0624 1.0610 1.0595 1.0580 log tan log cos 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9985 9.9985 9.9985 9.99S5 9.9985 9.9985 9.9985 9.9985 9.9985 9.9985 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9983 85° 60' 85° 59' 85° 58' 85° 57' 85° 56' 85° 55' 85° 54' 85° 53' 85° 52' 85° 51' 85° 50' 85° 49' 85° 48' 85° 47' 85° 46' 85° 45' 85° 44' 85° 43' 85° 42' 85° 41' 85° 40' 85° 39' 85° 38' 85° 37' 85° 36' 85° 35' 85° 34' 85° 33' 85° 32' 85° 31' 85° 30' 85° 29' 85° 28' 85° 27' 85° 26' 85° 25' 85° 24' 85° 23' 85° 22' 85° 21' 85° 20' 85° 19' 85° 18' 85° 17' 85° 16' 85° 15' 85° 14' 85° 13' 85° 12' 85° 11' 85° 10' 85° 9' 85° 85° 85° 85° 85° 85° 85° 85° 85° 7' 6' 5' 4' 3' 2' 1' 0' log sin Angle J COSINES, TANGENTS, AND COTANGENTS 13 5°-15° Angle log sin diff.l' log tan com. diff.l' log cot log cos diff.l' 5° 0' 5° 10' 8.9403 14.2 8.9420 14.3 1.0580 9.9983 .1 85° 0' 84° 50' 8.9545 8.9563 1.0437 9.9982 5° 20' 8.9682 13.7 8.9701 13.8 1.0299 9.9981 .1 84° 40' 5° 30' 8.9816 13.4 8.9836 13.5 1.0164 9.9980 .1 84° 30' 5° 40' 8.9945 12.9 8.9966 13.0 1.0034 9.9979 .1 84° 20' 5° SO' 6° 0' 6° 10' 9.0070 9.0192 12.5 12.2 11.9 9.0093 12.7 12.3 12.0 0.9907 9.9977 .2 .1 .1 84° 10' 84° 0' 83° 50' 9.0216 U.97S4 9.9976 9.0311 9.0336 0.9664 9.997S 6° 20' 9.0426 11.5 9.0453 11.7 0.9547 9.9973 .2 83° 40' 6° 30' 9.0539 11.3 9.0567 11.4 0.9433 9.9972 .1 83° 30' 6° 40' 9.0648 10.9 9.067S 11.1 0.9322 9.9971 .1 83° 20' 6° SO' 7° 0' 7° 10' 9.0755 10.7 10.4 10.2 9.0786 10.8 10.5 10.4 0.9214 9.9969 .2 .1 .2 83° 10' 83° 0' 82° SC 9.0S59 9.0891 0.9109 9.9968 9.0961 9.0995 0.9005 9.9966 7° 20' 9.1060 9.9 9.1096 10.1 0.S904 9.9964 .2 82° 40' 7° 30' 9.1157 9.7 9.1194 9.8 0.8806 9.9963 .1 82° 30' 7° 40' 9.12S2 9.5 9.1291 9.7 0.8709 9.9961 .2 82° 20' 7° SO' 8° 0' S° 10' 9.1345 9.3 9.1 8.9 9.138S 9.4 9.3 9.1 0.8615 9.9959 .2 .1 .2 82° 10' 82° 0' 81° 50' 9.1436 9.1478 0.8522 9.9958 9.1525 9.1569 0.8431 9.9956 8° 20' 9.1612 8.7 9.1658 8.9 0.8342 9.9954 .2 81° 40' 8° 30' 9.1697 8.5 9.1745 8.7 0.825S 9.9952 .2 81° 30' 8° 40' 9.1781 8.4 9.1831 8.6 0.8169 9.99S0 .2 81° 20' 8° SO' 9° 0' 9° 10' 9.1863 8.2 8.0 7.9 9.1915 8.4 8.2 8.1 0.S08S 9.9948 .2 .2 .2 81° 10' 81° 0' 80° 50' 9.1943 9.1997 0.8003 9.9946 9.2022 9.2078 0.7922 9.9944 9° 20' 9.2100 7.8 9.2158 8.0 0.7842 9.9942 .2 80° 40' 9° 30' 9.2176 7.6 9.2236 7.8 0.7764 9.9940 .2 80° 30' 9° 40' 9.22S1 7.5 9.2313 7.7 0.7687 9.9938 .2 80° 20' 9° 50' 10° 0' 10° 10' 9.2324 7.3 7.3 7.1 9.2389 7.6 7.4 7.3 0.7611 9.9936 .2 .2 .3 80° 10' 80° 0' 79° 50' 9.2397 9.2463 0.7537 9.9934 9.2468 9.2536 0.7464 9.9931 10° 20' 9.2538 7.0 9.2609 7.3 0.7391 9.9929 .2 79° 40' 10° 30' 9.2606 6.8 9.2680 7.1 0.7320 9.9927 .2 79° 30' 10° 40' 9.2674 6.8 9.2750 7.0 0.72SO 9.9924 79° 20' 10° 5C 11° 0' 11° 10' 9.2740 6.6 6.6 6.4 9.2S19 6.9 6.8 6.6 0.7181 9.9922 .2 .3 .2 79° 10' 79° 0' 78° SO' 9.2806 9.2887 0.7113 9.9919 9.2870 9.2953 0.7047 9.9917 11° 20' 9.2934 6.4 9.3020 6.7 0.6980 9.9914 .3 7S° 40' 11° 3<y 9.2997 6.3 9.3085 6.5 0.6915 9.9912 .2 78° 3C 11° 40' 9.3058 6.1 9.3149 6.4 0.68S1 9.9909 .3 78° 20' 11° SO' 12° 0' 12° 10' 9.3119 6.1 6.0 5.9 9.3212 6.3 6.3 6.1 0.6788 9.9907 .3 .3 78° IV 78° 0' 77° 50' 9.3179 9.3275 0.6725 9.9904 9.3238 9.3336 0.6664 9.9901 12° 20' 9.3296 5.8 9.3397 6.1 0.6603 9.9899 77° 40' 12° 30' 9.3353 5.7 9.3458 6.1 0.6542 9.9896 .3 77° 3C 12° 40' 9.3410 5.7 9.3S17 5.9 0.6483 9.9893 77° 20' 12° 50' 13° 0' 13° 1C 9.3466 5.6 5.5 5.4 9.3576 5.9 5.8 5.7 0.6424 9.9S90 .3 .3 .3 77° 10' 77° 0' 76° 50' 9.3521 9.3634 0.6366 9.9887 9.3S75 9.3691 0.6309 9.98S4 13° 20' 9.3629 5.4 9.3748 5.7 0.6252 9.9881 76° 40' 13° 30' 9.3682 5.3 9.3804 5.6 0.6196 9.9878 76° 30' 13° 4C 9.3734 5.2 9.3859 5.5 0.6141 9.9875 76° 20' 13° SO- 14° 0' 14° 10' 9.3786 5.2 5.1 5.0 9.3914 5.5 5.4 5.3 0.6086 9.9872 .3 .3 .3 .4 .3 76° 10' 76° 0' 75° SC 9.3837 9.3968 0.6032 9.9869 9.38S7 9.4021 0.5979 9.9866 14° 20' 9.3937 5.0 9.4074 5.3 0.5926 9.9863 75° 40' 14° 30' 9.3986 4.9 9.4127 5.3 0.5873 9.9859 75° 3(f 14° 40- 9.4035 4.9 9.4178 5.1 0.S822 9.9856 75° 20' 14° 50' 15° 0' 9.4083 4.8 4.7 9.4230 5.2 5.1 O.S770 9.98S3 .3 .4 75° 10' 75° 0' 9.4130 9.4281 0.5719 9.9849 log cos diff.l' log cot com. diff. 1' log tan log sin diff.l' Angle 75°-85° u TABLE II. LOGARITHMIC SINES 15°-25° Angle log sin cliff. 1' log tan com. difi.r log cot log cos cliff. 1' 15° 0' 15° 10' 9.4130 4.7 9.4281 9.4331 5.0 0.5719 9.9849 .3 75° 0' 74° 50' 9.4177 0.5669 9.9846 15° 20' 9.4223 4.6 9.4381 5.0 0.5619 9.9843 .3 74° 4C 15° 3V 9.4269 4.6 9.4430 4.9 4.9 0.5570 9.9839 .4 74° 30' 15° 40' 9.4314 9.4479 0.5521 9.9836 .3 74° 2V 15° SO' 16° 0' 16° 10' 9.4359 4.5 4.4 4.4 9.4527 4.8 4.8 4.7 0.5473 9.9832 .4 .4 .3 74° 10' 74° 0' 73° 50' 9.4403 9.4575 0.5425 9.9828 9.4447 9.4622 0.5378 9.9825 16° 20' 9.4491 4.4 9.4669 4.7 0.5331 9.9821 .4 73° 40' 16° 30' 9.4533 4.2 9.4716 4.7 0.5284 9.9817 .4 73° 30' 16° 40' 9.4576 4.3 9.4762 4.6 0.5238 9.9814 .3 73° 20' 16° SO' 17° 0' 17° 10' 9.4618 4.2 4.1 4.1 9.4808 4.6 4.5 4.5 0.5192 9.9810 .4 .4 .4 73° 10' 73° 0' 72° 5C 9.4659 9.4853 0.5147 9.9806 9.4700 9.4898 0.5102 9.9802 17° 20' 9.4741 4.1 9.4943 4.5 0.5057 9.9798 .4 72° 40- 17° 30' 9.4781 4.0 9.4987 4.4 0.5013 9.9794 .4 72° 30* 17° 40' 9.4821 4.0 9.5031 4.4 0.4969 9.9790 .4 72° 20' 17° 50' 18° 0' 18° 1C 9.4861 4.0 3.9 3.9 9.5075 4.4 4.3 4.3 0.4925 9.9786 .4 .4 .4 72° 10' 72° 0' 71° 50' 9.4900 9.5118 0.4882 9.9782 9.4939 9.5161 0.4839 9.9778 18° 20' 9.4977 3.8 9.S203 4.2 0.4797 9.9774 .4 71° 40' 18° 30' 9.5015 3.8 9.5245 4.2 0.4755 9.9770 .4 71° 30' 18° 40' 9.5052 3.7 9.5287 4.2 0.4713 9.9765 .5 71° 2C 18° 50' 19° 0' 19° 10' 9.5090 3.8 3.6 3.7 9.5329 4.2 4.1 4.1 0.4671 9.9761 .4 .4 .5 71° 10' 71° 0' 70° 50' 9.5126 9.5370 0.4630 9.9757 9.5163 9.5411 0.4589 9.9752 19° 20' 9.5199 3.6 9.5451 4.0 0.4549 9.9748 .4 70° 40' 19° 30' 9.5235 9.5491 4.0 0.4509 9.9743 .5 70° 30' 19° 40' 9.5270 9.5531 4.0 0.4469 9.9739 .4 70° 20' 19° 50' 20° 0' 20° IV 9.5306 3.6 3.5 3.4 9.5571 4.0 4.0 3.9 0.4429 9.9734 .5 .4 .5 70° 10' 70° 0' 69° 50' 9.5341 9.5611 0.4389 9.9730 9.5375 9.5650 0.4350 9.9725 20° 20' 9.5409 3.4 9.5689 3.9 0.4311 9.9721 .4 ! 69° 40' 20° 30' 9.5443 9.5727 3.8 0.4273 9.9716 .5 1 69° 30' 20° 40' 9.5477 3.3 3.3 3.3 9.5766 3.9 0.4234 9.9711 .5 69° 20' 20° SO' 21° 0' 21° 10' 9.5510 9.5804 3.8 3.8 3.7 0.4196 9.9706 .5 .4 .5 69° 10' 69° 0' 68° 50' 9.5543 9.5842 0.4158 9.9702 9.5576 9.5879 0.4121 9.9697 21° 20' 9.5609 9.5917 3.8 0.4083 9.9692 .5 68° 40' 21° 30' 9.5641 3.2 9.5954 3.7 0.4046 9.9687 .5 68° 30' 21° 40' 9.5673 3.2 9.5991 3.7 0.4009 9.9682 .5 68° 20' 21° 50' 22° 0' 22° 1C 9.5704 3.1 3.2 3.1 9.6028 3.7 3.6 3.6 0.3972 9.9677 .5 .5 .5 68° 10' 68° 0' 67° 50' 9.5736 9.6064 0.3936 9.9672 9.5767 9.6100 0.3900 9.9667 22° 20' 9.S798 3.1 9.6136 3.6 0.3864 9.9661 .6 67° 40' 22° 30' 9.S828 3.0 9.6172 3.6 0.3828 9.9656 .5 67° 30' 22° 40' 9.5859 .3.1 9.6208 3.6 0.3792 9.9651 .5 67° 20' 22° 50' 23° 0' 23° 10' 9.5889 3.0 3.0 2.9 9.6243 3.5 3.6 3.5 0.3757 9.9646 .5 .6 .5 67° 10' 67° 0' 66° 50' 9.5919 9.6279 0.3721 9.9640 9.5948 9.6314 0.3686 9.9635 23° 20' 9.5978 3.0 9.6348 3.4 0.3652 9.9629 .6 ! 66° 40' 23° 30' 9.6007 2.9 9.6383 3.5 0.3617 9.9624 .5 66° 30' 23° W 9.6036 9.6417 3.4 0.3583 9.9618 .6 66° 20' 23° 50' 24° 0' 24° 10' 9.6065 2.9 2.8 2.8 9.6452 3.5 3.4 3.4 0.3548 9.9613 .5 .6 .5 66° 10- 66° 0' 65° 5C 9.6093 9.6486 0.3514 9.9607 9.6121 9.6520 0.3480 9.9602 24° 20" 9.6149 2.8 9.6553 3.3 0.3447 9.9596 .6 65° W 24° 30' 9.6177 2.8 9.6587 3.4 0.3413 9.9590 .6 65° 3C 24° 40' 9.6205 2.8 9.6620 3.3 0.3380 9.9584 .6 65° 2C 24° 50' 25° 0' 9.6232 2.7 2.7 9.6654 3.4 3.3 0.3346 9.9579 .5 .6 65° 10' 65° 0' 9.6259 9.6687 0.3313 9.9573 log cos diff. 1' log cot com. die. V log tan log sin diff. 1' Angle 6 >5°-75 COSINES, TANGENTS, AND COTANGENTS 15 25°-3S 1° Angle log sin diff.1' log tan com. diff.r log cot log cos dij 1.1' 25° 0' 25° 10' 9.6259 2.7 9.6687 3.3 0.3313 9.9573 6 65° 0' 64° 5C 9.6286 9.6720 0.3280 9.9567 25° 20 / 9.6313 2.7 9.6752 3.2 " 0.3248 9.9561 6 64° 40' 25° 30' 9.6340 2.7 9.6785 3.3 0.3215 9.9555 6 64° 30' 25° 40' 9.6366 2.6 9.6817' 3.2 0.3183 9.9549 6 64° 20' 25° 50" 26° 0' 26° 1C 9.6392 2.6 2.6 2.6 9.6850 3.3 3.2 3.2 0.3150 9.9543 6 6 7 64° 10' 64° 0' 63° 50' 9.6418 9.6882 0.3118 9.9537 9.6444 9.6914 0.3086 9.9530 26° 20' 9.6470 2.6 9.6946 3.2 0.3054 9.9524 6 63° 40" 26° 30' 9.6495 2.5 9.6977 3.1 0.3023 9.9518 6 63° 30' 26° 40' 9.6521 2.6 9.7009 3.2 0.2991 9.9512 6 63° 20- 26° SW 27° 0' 27° 10' 9.6546 2.5 2.4 2.5 9.7040 3.1 3.2 3.1 0.2960 9.9505 7 6 7 63° 10' 63° 0' 62° 50' 9.6570 9.7072 0.2928 9.9499 9.6595 9.7103 0.2897 9.9492 27° 20' 9.6620 2.5 9.7134 3.1 0.2866 9.9486 6 62° 40' 27° 3^ 9.6644 2.4 9.7165 3.1 0.2835 9.9479 7 62° 30' 27° 4C 9.6668 2.4 9.7196 3.1 0.2804 9.9473 6 62° 20' 27° 5V 28° 0' 28° 10' 9.6692 2.4 2.4 2.4 9.7226 3.0 3.1 3.0 0.2774 9.9466 7 7 6 62° 10 7 62° 0' 61° 50' 9.6716 9.7257 0.2743 9.9459 9.6740 9.7287 0.2713 9.9453 28° 20' 9.6763 2.3 9.7317 3.0 0.2683 9.9446 7 61° W 28° 3CK 9.6787 2.4 9.7348 3.1 0.2652 9.9439 7 61° 3^ 28° 4C 9.6810 2.3 9.7378 3.0 0.2622 9.9432 7 61° 20' 28° 5C 29° 0' 29° 1C 9.6833 2.3 2.3 2.2 9.7408 3.0 3.0 2.9 0.2592 9.9425 7 7 7 61° 1C 61° C 60° 50" 9.6856 9.7438 0.2562 9.9418 9.6878 9.7467 0.2533 9.9411 29° 20' 9.6901 2.3 9.7497 3.0 0.2503 9.9404 7 60° 40' 29° 30' 9.6923 2.2 9.7526 2.9 0.2474 9.9397 7 60° 30" 29° W 9.6946 2.3 9.7556 3.0 0.2444 9.9390 7 60° 20' 1<P 50" 30° 0' 30° 1C 9.6968 2.2 2.2 2.2 9.7585 2.9 2.9 2.9 0.2415 9.9383 7 8 7 60° 10' 60° C 59° 5C 9.6990 9.7614 0.2386 9.9375 9.7012 9.7644 0.2356 9.9368 30° 20' 9.7033 2.1 9.7673 2.9 0.2327 9.9361 7 59° 40- 30° 30' 9.7055 2.2 9.7701 2.8 0.2299 9.9353 8 59° 3C 30° 40' 9.7076 2.1 9.7730 2.9 0.2270 9.9346 7 59° 2C 30° 50' 31° 0' 31° 10' 9.7097 2.1 2.1 2.1 9.7759 2.9 2.9 2.8 0.2241 9.9338 8 .7 .8 59° 1C 59° W 58° 5C 9.7118 9.7788 0.2212 9.9331 9.7139 9.7816 0.2184 9.9323 3i° 2<y 9.7160 2.1 9.7845 2.9 0.2155 9.9315 8 58° 40' 31° 3C 9.7181 2.1 9.7873 2.8 0.2127 9.9308 7 58° 30' 31° 40 7 9.7201 2.0 9.7902 2.9 0.2098 9.9300 .8 58° 20' 31° 50' 32° 0' 32° 10' 9.7222 9.7242 9.7262 2.1 2.0 2.0 9.7930 2.8 2.8 2.8 0.2070 9.9292 8 8 8 58° 10- 58° 0' 57° 5C 9.7958 0.2042 9.9284 9.7986 0.2014 9.9276 32° 2(y 9.7282 2.0 9.8014 2.8 0.1986 9.9268 8 57° 40' 32° 3<y 9.7302 2.0 9.8042 2.8 0.1958 9.9260 8 57° 30' 32° 4C 9.7322 2.0 9.8070 2.8 0.1930 9.9252 8 57° 20" 32° 5C 33° 0' 33° 10- 9.7342 2.0 1.9 1.9 9.8097 2.7 2.8 2.8 0.1903 9.9244 8 8 8 9 8 8 9 8 9 8 9 9 57° 1C 57° C 56° 5C 9.7361 9.8125 0.1875 9.9236 9.7380 9.8153 0.1847 9.9228 33° 2W 9.7400 2.0 9.8180 2.7 0.1820 9.9219 56° 40' 33° 30' 9.7419 1.9 9.8208 2.8 0.1792 9.9211 56° 30" 33° 40' 9.7438 1.9 9.8235 2.7 0.1765 9.9203 56° 20' 33° sor 34° 0' 34° IV 9.7457 1.9 1.9 1.8 9.8263 2.8 2.7 2.7 0.1737 9.9194 56° 1C 56° 0' 55° sty 9.7476 9.8290 0.1710 9.9186 9.7494 9.8317 0.1683 9.9177 34° 20- 9.7513 1.9 9.8344 2-7 0.1656 9.9169 55° 4C 34° 3C 9.7531 1.8 9.8371 2 7 0.1629 9.9160 55° 3C 34 e W 9.7550 1.9 9.8398 2.7 0.1602 9.9151 55° 20' 34° 50- 35° 0' 9.7568 1.8 1.8 9.8425 2.7 2.7 0.1575 9.9142 9 8 55° 1C 55° 0' 9.7586 9.8452 0.1548 9.9134 log cos diff.1' log cot com. diff.1' log tan log sin dil 5.1' Angle 55°-65 1° 16 TABLE II. LOGARITHMIC SINES 35°- 45° Angle log sin difi. 1 l°gtan ffii log cot log COS difi. 1' 35° 0' 35° 10' 35° 2C 35° 30' 35° W 35° SO' 36° 0' 36° 10' 36° 20' 36° 30' 36° 40' 36° SO' 37° 0' 37° 10' 37° 20' 37° 30' 37° W 37° SO- 38° 0' 38° 10' 38° 20' 38° 30' 38° 40' 38° 50' 39° 0' 39° i<y 39° 20' 39° 30' 39° 40' 39° SO' 40° 0' 40° IV 40° 20' 40° 30' 40° 4C 40° SV 41° 0' 41° 10' 41° 20' 41° 30' 41° 40' 41° SO' 42° 0' 42° 10' 42° 20' 42° 30' 42° 40' 42° SO' 43° 0' 43° 10' 43° 20' 43° 30' 43° 40' 43° SO' 44° 0' 44° 10' 44° 20' 44° 30' 44° 40' 44° SO' 45° 0' 9.7S86 9.7604 9.7622 9.7640 9.76S7 9.7675 9.7692 9.7710 9.7727 9.7744 9.7761 9.7778 9. 7795 9.7811 9.7828 9.7844 9.7861 9.7877 9.7893 9.7910 9.7926 9.7941 9.7957 9.7973 9.7989 9.8004 9.8020 9.8035 9.8050 9.8066 J.S081 9.8096 9.8111 9.812S 9.8140 9.815S 9.8169 9.8184 9.8198 9.8213 9.8227 9.8241 9.S255 9.8269 9.8283 9.8297 9.8311 9.8324 9.8338 9.8351 9.8365 9.8378 9.8391 9.8405 9.8418 9.8431 9.8444 9.8457 9.8469 9.8482 9.8495 log cos difi. 1 1.8 1.8 1.8 1.7 1.8 1.7 1.8 1.7 1.7 1.7 1.7 1.7 1.6 1.7 1.6 1.7 1.6 1.6 1.7 1.6 1.5 1.6 1.6 1.6 1.5 1.6 1.5 1.5 1.6 1.5 1.5 1.5 1.4 1.5 1.5 1.4 1.5 1.4 1.5 1.4 1.4 1.4 1.4 1.4 1.4 1.4 1.3 1.4 1.3 1.4 1.3 1.3 1.4 1.3 1.3 1.3 1.3 1.2 1.3 1.3 9.8452 9.8479 9.8506 9.8533 9.8S59 9.8586 9.8613 9.8639 9.8666 9.8692 9.8718 9.874S 9.8771 9.8797 9.8824 9.8850 9.8876 9.8902 9.8928 9.8954 9.8980 9.9006 9.9032 9.905 S 9.9084 9.9110 9.9135 9.9161 9.9187 9.9212 9.9238 9.9264 9.9289 9.9315 9.9341 9.9366 9.9392 9.9417 9.9443 9.9468 9.9494 9.9519 9.9544 9.9570 9.9S95 9.9621 9.9646 9.9671 9.9697 9.9722 9.9747 9.9772 9.9798 9.9823 9.9848 9.9874 9.9899 9.9924 9.9949 9.9975 0.0000 2.7 2.7 2.7 2.6 2.7 2.7 2.6 2.7 2.6 2.6 2.7 2.6 2.6 2.7 2.6 2.6 2.6 2.6 2.6 2.6 2.6 2.6 2.6 2.6 2.6 2.5 2.6 2.6 2.5 2.6 2.6 2.5 2.6 2.6 2.5 2.6 2.5 2.6 2.5 2.6 2.5 2.5 2.6 2.5 2.6 2.5 2.5 2.6 2.5 2.5 2.5 2.6 2.5 2.5 2.6 2.5 2.5 2.5 2.6 2.5 0.1S48 9.9134 0.1S21 0.1494 0.1467 0.1441 0.1414 9.912S 9.9116 9.9107 9.9098 9.9089 0.1387 0.1361 0.1334 0.1308 0.1282 0.1255 0.1229 0.1203 0.1176 0.1150 0.1124 0.1098 0.1072 0.1046 0.1020 0.0994 0.0968 0.0942 0.0916 0.0890 0.0865 0.0839 0.0813 0.0788 0.0762 0.0736 0.0711 0.0685 0.0659 0.0634 0.0608 0.0583 0.0SS7 0.0532 0.0506 0.0481 0.04S6 0.0430 0.0405 0.0379 0.03S4 0.0329 0.0303 0.0278 0.02S3 0.0228 0.0202 0.0177 0.0152 0.0126 0.0101 0.0076 0.0051 0.0025 0.0000 log cot ^i, log tan 45°- 55° 9.9080 9.9070 9.9061 9.9052 9.9042 9.9033 9.9023 9.9014 9.9004 9.899S 9.898S 9.8975 9.8965 9.8955 9.8945 9.8935 9.8925 9.8915 9.8905 9.889S 9.8884 9.8874 9.8864 9.8853 9.8843 9.8832 9.8821 9.8810 9.8800 9.8789 9.8778 9.8767 9.8756 9.874S 9.8733 9.8722 9.8711 9.8699 9.8688 9.8676 9.8665 9.8653 9.8641 9.8629 9.8618 9.8606 9.8594 9.8582 9.8569 9.85S7 9.8S4S 9.8S32 9.8520 9.8S07 9.8495 .9 .9 .9 .9 .9 .9 1.0 .9 .9 1.0 .9 1.0 .9 1.0 .9 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.1 1.0 1.0 1.1 1.0 1.1 1.1 1.1 1.0 1.1 1.1 1.1 l.l 1.1 1.2 1.1 1.1 1.2 1.1 1.2 1.1 1.2 1.2 1.2 1.1 1.2 1.2 1.2 1.3 1.2 1.2 1.3 1.2 1.3 1.2 log sin difi. V 55° 0' 54° 50' 54° 40' 54° 30' 54° 20' 54° 10' 54° 0' 53° SO' 53° 40' 53° 30' 53° 20' 53° 10' 53° 0' 52° SO' 52° 40' 52° 30' 52° 20' 52° 10' 52° 0' 51° 50' 51° 40' 51° 30' 51° 20' 51° 1C 51° 0' 50° SO' 50° 40' 50° 30' 50° 20' 50° 10' 50° 0' 49° 50' 49° 40' 49° 30' 49° 20' 49° 10' 49° 0' 48° SO' 48° 40' 48° 30' 48° 20' 48° 10' 48° 0' 47° 50' 47° 40' 47° 30' 47° 20' 47° 10' 47° 0' 46° 50' 46° 40' 46° 30' 46° 20' 46° 10' 46° 0' 45° 50' 45° 40' 45° 30' 45° 20' 45° 10' 45° 0' Angle CONVERSION TABLES FOR ANGLES 17 To change fkom Minutes and Seconds into the Decimal Parts of a Degree or into Radians From seconds 1"= 2"= 3"= 4"= 5" = 6"= 7"= 8"= 9"= 10"= 20"= 30"= 40"= 50"= 0.00028°= 0.00056° = 0.00083°= 0.00111°= 0.00139°= 0.00167°= 0.00194°= 0.00222°= 0.00250°= 0.00278° = 0.00556°= 0.00833°= 0.01111°= 0.01389°= 0.0000048 Rad. 0.0000097 0.0000145 : 0.0000194 : 0.0000242 ; 0.0000291 0.0000339 : 0.0000388 0.0000436 0.0000485 0.0000970 0.0001454 0.0001939 0.0002424 From minutes l'=0.017°= 2'= 0.033°= 3'=0.050°= 4' =0.067°= 5' =0.083°= 6'=0.100°= 7'=0.117°= 8'=0.133°= 9'=0.150°= 10'=0.167°= 20'=0.333°= 30'=0.500°= 40' =0.667°= 50'=0.833°= 0.00029 Rad. :0.00058 " 0.00087 " : 0.00116 " : 0.00145 " : 0.00175 " 0.00204 " :0.00233 " 0.00262 " 0.00291 " 0.00582 " 0.00873 " 0.01164 " 0.01454 " From degrees into radians 1°=0 2°=0. 3°=0 4°=0 5°=0 6°=0 7°=0 8°=0 9°=0 10° =0. 20° =0. 30° =0. 40° =0. 50° =0. .01745 Rad. 03491 .05236 .06981 .08727 .10472 .12217 13963 .15708 17453 34907 52360 69813 87266 to change from decimal parts of a degree into mlnutes and Seconds 0.0000° = 0.000' = 0" 0.20° = 12.0' = 12' 0.60° = 36.0' = 36' 0.0001° = 0.006' = 0.36" 0.21° = 12.6' = 12' 36" 0.61° = 36.6' = 36' 36" 0.0002° = 0.012' = 0.72" 0.22° = 13.2' = 13' 12" 0.62° = 37.2' = 37' 12" 0.0003° = 0.018' = 1.08" 0.23° = 13.8' = 13' 48" 0.63° = 37.8' = 37' 48" 0.0004° = 0.024' = 1.44" 0.24° = 14.4' = 14' 24" 0.64° = 38.4' = 38' 24" 0.0005° = 0.030' = 1.80" 0.25° = 15.0' = 15' 0.65° = 39.0' = 39' 0.0006° = 0.036' = 2.16" 0.26° = 15.6' = 15' 36" 0.66° = 39.6' = 39' 36" 0.0007° = 0.042' = 2.52" 0.27° = 16.2' = 16' 12" 0.67° = 40.2' = 40' 12" 0.0008° = 0.048' = 2.88" 0.28° = 16.8' = 16' 48" 0.68° = 40.8' = 40' 48" 0.0009° = 0.054' = 3.24" 0.29° = 17.4' = 17' 24" 0.69° = 41.4' = 41' 24" 0.0010° = 0.06(r = 3.60" 0.30° = 18.0' = 18' 0.70° = 42.0' = 42' 0.001° = 0.06' = 3.6" 0.31° = 18.6' = 18' 36" 0.71° = 42.6' = 42' 36" 0.002° = 0.12' = 7.2" 0.32° = 19.2' = V¥ 12" 0.72° = 43.2' = 43' 12" 0-003° = 0.18' = 10.8" 0.33° = 19.8' = 19' 48" 0.73° = 43.8' = 43' 48" 0.004° = 0.24' = 14.4" 0.34° = 20.4' = 20' 24" 0.74° = 44.4' = 44' 24" 0.005° = 0.30' = 18.0" 0.35° = 21.0' = 21' 0.75° = 45.0' = 45' 0.006° = 0.36' = 21.6" 0.36° = 21.6' = 21' 36" 0.76° = 45.6' = 45' 36" 0.007° = 0.42' = 25.2" 0.37° = 22.2' = 22' 12" 0.77° = 46.2' = 46' 12" 0.008° = 0.48' = 28.8" 0.38° = 22.8' = 22' 48" 0.78° = 46.8' = 46' 48" 0.009° = 0.54' = 32.4" 0.39° = 23.4' = 23' 24" 0.79° = 47.4' = 47' 24" 0.010° = 0.60' = 36.0" 0.40° = 24.0' = 24' 0.80° = 48.0' = 48' 0.01° = 0.6' = 36" 0.41° = 24.6' = 24' 36" 0.81° = 48.6' = 48' 36" 0.02° = 1.2' = 1' 12" 0.42° = 25.2' = 25' 12" O.S2° = 49.2' = 49' 12" 0.03° = 1.8' = 1' 48" 0.43° = 25.8' = 25' 48" 0.83° = 49.8' = 49' 48" 0.04° = 2.4' = 2' 24" 0.44° = 26.4' = 26' 24" 0.84° = 50.4' = 50' 24" 0.05° = 3.0' = 3' 0.45° = 27.0' = 27' 0.85° = 51.0' = 51' 0.06° = 3.6' = 3' 36" 0.46° = 27.6' = 27' 36" 0.86° = 51.6' = 51' 36" 0.07° ' = 4.2' = 4' 12" 0.47° = 28.2' = 28' 12" 0.87° = 52.2' = 52' 12" 0.08° = 4.8' = 4' 48" 0.48° = 28.8' = 28' 48" 0.88° = 52.8' = 52' 48" 0.09° = 5.4' = 5' 24" 0.49° = 29.4' = 29' 24" 0.89° = 53.4' = 53' 24" 0.10° = 6.0' = 6' 0.50° = 30.0' = 30' 0.90° = 54.0' = 54' 0.11° = 6.6' = 6' 36" 0.51° = 30.6' = 30' 36" 0.91° = 54.6' = 54' 36" 0.12° = 7.2' = 7' 12" 0.52° = 31.2' = 31' 12" 0.92° = 55.2' = 55' 12" 0.13° = 7.8' = 7' 48" 0.53° = 31.8' = 31' 48" 0.93° = 55.8' = 55' 48" 0.14° = 8.4' = 8' 24" 0.54° = 32.4' = 32' 24" 0.94° = 56.4' = 56' 24" 0.15° = 9.V = 9' 0.55° = 33.C = 33' 0.95° = 57.C = 57' 0.16° = 9.6' = 9'36" 0.56° = 33.6' = 33' 36" 0.96° = 57.6' = 57' 36" 0.17° = 10.2' = 10' 12" 0.57° = 34.2' = 34' 12" 0.97° = 58.2' = 58' 12" 0.18° = 10.8' = 10' 48" 0.58° = 34.8' = 34' 48" 0.98° = 58.8' = 58' 48" 0.19° = 11.4' = 11' 24" 0.59° = 35.4' = 35' 24" 0.99° = 59.4' = 59' 24" 0.20° = 12.CC = 12' 0.60° = 36.0' = 36' 1.00° = 60.0' = 60' Table III FOUR-PLACE LOGARITHMS OF TRIGONOMETRIC FUNCTIONS, THE ANGLE BEING EXPRESSED IN DEGREES AND THE DECIMAL PART OF A DEGREE This table gives the common logarithms (base 1(T) of the sines, cosines, tangents, and cotangents of all angles from 0° to 5°, and from 85° to 90° for every hundredth part of a degree, and from 5° to 85° for every tenth of a degree, all calculated to four places of decimals. In order to avoid the printing of negative characteristics, the number 10 has been added to every logarithm in the first, second, and fourth columns ^those having log sin, log tan, and log cos at the top). Hence in writing down any logarithm taken from these three columns — 10 should be written after it. Loga- rithms taken from the third column (having log cot at the top) should be used as printed. A logarithm found from this table by interpolation may be in error by one unit in the last decimal place, except for angles between 0° and 0.3° or between 89.7° and 90°, when the error may be larger. In the latter cases the table refers the student to the formulas on page 6 for more accurate results. 19 20 TABLE III. LOGARITHMIC SINES 0° Angle log sin diS. log tan com. diK. log cot log cos Prop. Parts 0.00° 0.01° . 10.0000 90.00° 89.99° an 6.2419 6.2419 3.7581 10.0000 0.02° 6.5429 6.5429 3.4571 10.0000 89.98° ti Difference 0.03° 6.7190 6.7190 3.2810 10.0000 89.97° 0.04° 6.8439 45 6.8439 to 3.1561 10.0000 89.96° H 0.05° 0.06° 6.9408 7.0200 3 m ? 6.9408 7.0200 CO u 3.0592 2.9800 10.0000 10.0000 89.95° 89.94° — 79 78 77 0.07° 7.0870 7.0870 9i 2.9130 10.0000 89.93° 1 9 7.9 15 8 7.8 15.6 7.7 15 4 0.08° 7.1450 d 7.1450 d 2.8550 10.0000 89.92° 3 23.7 23.4 23.1 0.09° 0.10° 7.1961 7.1961 .9 2.8039 10.0000 89.91° 89.90° 4 5 6 31.6 39.5 47.4 31.2 39.0 46.8 30.8 38.5 46.2 7.2419 7.2419 2.7581 10.0000 0.11° 7.2833 > 7.2833 'Ed 2.7167 10.0000 89.89° l 55.3 63.2 54.6 62.4 53.9 61.6 0.12° 7.3211 7.3211 I— 1 2.6789 10.0000 89.88° 9 71.1 70.2 69.3 0.13° 7.3558 u a 7.3558 t-i 2.6442 10.0000 89.87° 76 75 74 0.14° 7.3880 bo 7.3880 ttD 2.6120 10.0000 89.86° 1 7.6 7.5 7.4 0.15° 7.4180 £ 7.4180 a 2.5820 10.0000 89.85° 2 15.2 15.0 14.8 0.16° 7.4460 pj 7.4460 s 2.5540 10.0000 89.84° 3 22.8 22.5 22.2 4 30.4 30.0 29.6 0.17° 7.4723 7.4723 2.5277 10.0000 89.83° 5 38.0 37.5 37.0 0.18° 7.4971 ■J — 7.4972 3) ft 2.5028 10.0000 89.82° b 7 45.6 53 2 45.0 52.5 44.4 51.8 0.19° 7.5206 a o 7.5206 2.4794 10.0000 89.81° 8 60.8 60.0 69.2 0.20° 0.21° 7.5429 33 3. S 7.5429 .3 3 8. a 2.4571 10.0000 89.80° 89.79° 9 68.4 73 67.5 72 66.6 71 7.5641 7.5641 2.4359 10.0000 0.22° 7.5843 7.5843 SS 2.4157 10.0000 89.78° 1 7.3 7.2 7.1 0.23° 7.6036 7.6036 4J ' H 2.3964 10.0000 89.77° 2 14.6 14.4 14.2 •H co it 21.9 21.6 21.3 0.24° 7.6221 7.6221 2.3779 10.0000 89.76° 4 29.2 28.8 28.4 0.25° 7.6398 .9 « 7.6398 2.3602 10.0000 89.75° i> 6 36.5 43.8 36.0 43.2 35.5 42.6 0.26° 7.6568 "d w 7.6569 "t3 to 2.3431 10.0000 89.74° 7 51.1 50.4 49.7 O fl 8 58.4 57.6 56.8 0.27° 0.28° 7.6732 7.6890 7.6732 7.6890 2.3268 2.3110 10.0000 10.0000 89.73° 89.72° 9 65.7 64.8 03.9 0.29° 0.30° 0.31° 7.7043 142 7.7043 7.7190 142 2.2957 10.0000 89.71° 89.70° 89.69° 1 2 3 69 6.9 13.8 20.7 68 6.8 13.6 20.4 67 6.7 13.4 20.1 7.7190 2.2810 10.0000 7.7332 7.7332 2.2668 10.0000 0.32° 7.7470 138 7.7470 138 2.2530 10.0000 89.6S° 4 27.6 34.5 27.2 34.0 26.8 33.5 0.33° 7.7604 134 7.7604 134 2.2396 10.0000 S9.67° 6 41 4 40.8 40.2 130 130 7 483 47.6 46.9 0.34° 7.7734 7.7734 2.2266 10.0000 89.66° 8 55.2 54.4 53.6 0.35° 7.7859 125 7.7860 126 2.2140 10.0000 89.65° 9 62.1 61.2 60.3 0.36° 7.7982 123 119 7.7982 122 119 2.2018 10.0000 89.64° 66 65 64 0.37° 7.8101 7.8101 2.1899 10.0000 89.63° 1 6.6 6.5 6.4 0.38° 7.8217 116 7.8217 116 2.1783 10.0000 89.62° 2 13.2 13.0 12.8 0.39° 0.40° 7.8329 112 110 7.8329 112 110 2.1671 10.0000 89.61° 89.60° 4 5 19.8 26.4 33.0 19.5 26.0 32.5 19.2 25.6 32.0 7.8439 7.8439 2.1561 10.0000 108 108 6 39 6 39 38.4 0.41° 7.8547 7.8547 2.1453 10.0000 89.59° 7 46.2 45.5 44.8 0.42° 7.8651 104 7.8651 104 2.1349 10.0000 89.58° 8 52.8 52.0 51.2 0.43° 7.8753 102 100 7.8754 103 99 2.1246 10.0000 89.57° 9 59.4 58.5 57.6 0.44° 7.8853 7.8853 2.1147 10.0000 89.56° 63 62 61 0.45° 7.8951 98 7.8951 98 2.1049 10.0000 89.55° 1 6.3 6.2 6.1 0.46° 7.9046 95 7.9046 95 2.0954 10.0000 89.54° 9 it 12.6 18.9 12.4 186 18.3 0.47° 7.9140 94 7.9140 94 2.0860 10.0000 89.53° 4 5 25.2 31.5 24.8 31.0 24.4 30.5 0.48° 7.9231 91 7.9231 91 2.0769 10.0000 S9.52° 6 37.8 37.2 36.6 0.49° 0.50° 7.9321 90 87 7.9321 90 88 2.0678 10.0000 89.51° 89.50° 7 8 9 44.1 50.4 56.7 43.4 49.6 55.8 42.7 48.8 54.9 7.9408 7.9409 2.0591 10.0000 log cos diS. log cot com. diK. log tan log sin Angle 89° COSINES, TANGENTS, AND COTANGENTS 21 0° Angle log sin dig. log tan com. diH. log cot log cos Prop. Parts 0.50° 0.51° 7.9408 86 7.9409 86 2.0591 10.0000 89.50° 89.49° 'So 7.9494 7.9495 2.0505 10.0000 0.52° 7.9579 85 82 82 7.9579 84 83 81 2.0421 10.0000 89.48° t3 Difference 0.53° 7.9661 7.9662 2.0338 10.0000 89.47° "5 0.54° 7.9743 7.9743 2.0257 10.0000 89.46° w 0.55° 0.56° 7.9822 7.9901 79 79 76 7.9823 7.9901 80 78 77 2.0177 2.0099 10.0000 10.0000 89.45° 89.44° 60 59 58 0.57° 7.9977 76 7.9978 2.0022 10.0000 89.43° i 9 6.0 12 5.9 11 8 5.8 11 6 0.58° 8.0053 8.0053 1.9947 10.0000 89.42° 3 18.0 17.7 17.4 0.59° 8.0127 74 8.0127 74 1.9873 10.0000 89.41° 4 24.0 23.6 23.2 73 73 5 30 29 5 29 0.60° 0.61° 8.0200 72 71 8.0200 72 1.9800 10.0000 89.40° 89.39° 6 7 8 36.0 42.0 4S.0 35.4 41.3 47.2 34.8 40.6 46 4 8.0272 8.0272 1.9728 10.0000 0.62° 8.0343 8.0343 1.9657 10.0000 89.38° 9 54.0 53.1 52.2 0.63° 8.0412 69 68 8.0412 69 69 1.9588 10.0000 89.37° 57 56 55 0.64° 8.0480 8.0481 1.9519 10.0000 89.36° 1 57 5.6 55 0.65° 8.0548 8.0548 67 1.9452 10.0000 89.35° 2 11.4 11.2 11.0 0.66° S.0614 66 8.0614 66 1.9386 10.0000 89.34° 3 17.1 16.8 16.5 65 66 4 22.8 22.4 22.0 0.67° 8.0679 8.0680 1.9320 10.0000 89.33° 5 28.5 28.0 27.5 0.68° 8.0744 65 8.0744 64 1.9256 10.0000 89.32° 6 7 34.2 39.9 33.6 39.2 33.0 38.5 0.69° 0.70° 0.71° 8.0807 63 63 61 8.0807 63 63 62 1.9193 10.0000 89.31° 89.30° 89.29° 8 9 45.6 51.3 44.8 50.4 44.0 49.5 8.0870 8.0870 1.9130 10.0000 54 53 52 8.0931 8.0932 1.9068 10.0000 0.72° 8.0992 61 8.0992 60 1.9008 10.0000 89.28° 1 5.4 5.3 5.2 0.73° 8.1052 60 59 8.1052 60 59 1.S948 10.0000 89.27° 2 3 10.8 16.2 10.6 15.9 10.4 15.6 0.74° 8.1111 8.1111 1.8889 10.0000 89.26° 4 21.6 21.2 20.8 0.75° 8.1169 58 8.1170 59 1.8830 10.0000 89.25° 5 fi 27.0 32.4 26.5 31.8 26.0 31.2 0.76° 8.1227 8.1227 1.8773 10.0000 89.24° 7 37.8 37.1 36.4 57 57 8 43.2 42 4 41.6 0.77° 8.1284 8.1284 1.8716 10.0000 89.23° 9 48.6 47.7 46.8 0.78° 8.1340 56 8.1340 56 1.8660 10.0000 89.22° 0.79° 0.80° 0.81° 8.1395 55 55 53 8.1395 55 55 54 1.8605 10.0000 89.21° 89.20° 89.19° 1 2 3 4 R 51 5.1 10.2 15.3 20.4 25.5 50 5.0 10.0 15.0 20.0 25.0 49 4.9 9.8 14.7 19.6 24.5 8.1450 8.1450 1.8550 10.0000 8.1503 8.1504 1.S496 10.0000 0.82° 8.1557 54 8.1557 53 1.8443 10.0000 89.18° 0.83° 8.1609 52 8.1610 1.8390 10.0000 89.17° 6 30.6 30.0 29.4 52 52 7 35.7 35.0 34.3 0.84° 8.1661 8.1662 1.8338 10.0000 89.16°; 8 40.8 40.0 39.2 0.85° 8.1713 52 8.1713 51 1.8287 10.0000 89.15° 9 45.9 .45.0 44.1 0.86° 8.1764 51 50 8.1764 51 50 1.8236 10.0000 89.14° 48 47 46 0.87° 8.1814 8.1814 1.8186 9.9999 89.13° 1 4.8 4.7 4.6 0.88° 8.1863 49 8.1864 50 1.8136 9.9999 89.12° 2 3 4 5 6 7 9.6 14.4 19.2 24.0 28.8 33.6 9.4 14.1 18.8 23.5 28.2 32.9 9.2 13:8 18.4 23.0 27.6 32.2 0.89° 0.90° 0.91° 8.1912 49 49 48 8.1913 49 49 48 1.8087 9.9999 89.11° 89.10° 89.09° 8.1961 8.1962 1.8038 9.9999 8.2009 8.2010 1.7990 9.9999 0.92° 8.2056 47 8.2057 47 1.7943 9.9999 89.08° 8 9 38.4 43.2 37.6 42.3 36.8 0.93° 8.2103 47 47 8.2104 47 46 1.7896 9.9999 89.07° 0.94° 8.2150 8.2150 1.7850 9.9999 89.06° 45 44 43 0.95° 8.2196 46 8.2196 46 1.7804 9.9999 89.05° 1 2 3 4.5 9.0 13 5 4.4 8.8 13 2 4.3 8.6 12.9 0.96° 8.2241 45 8.2242 46 1.7758 9.9999 89.04° 0.97° 8.2286 8.2287 1.7713 9.9999 89.03° 4 5 18.0 W.,5 17.6 22,0 17.2 21.5 0.98° 8.2331 45 8.2331 44 1.7669 99999 89.02° 6 27.0 26.4 25.8 0.99° 1.00° 8.2375 44 44 8.2376 45 43 1.7624 9.9999 89.01° 89.00° 7 8 9 31.5 36.0 40.5 30.8 35.2 39.6 30.1 34.4 38.7 8.2419 8.2419 1.7581 9.9999 log cos diH. log cot com. diH. log tan log sin Angle 89° 22 TABLE III. LOGAKITHMIC SINES Angle 1.00° 1.01° 1.02° 1.03° 1.04° 1.05° 1.06° 1.07° 1.08° 1.09° 1.10° 1.11° 1.12° 1.13° 1.14° 1.15° 1.16° 1.17° 1.18° 1.19° 1.20° 1.21° 1.22° 1.23° 1.24° 1.25° 1.26° 1.27° 1.28° 1.29° 1.30° 1.31° 1.32° 1.33° 1.34° 1.35° 1.36° 1.37° .1.38° 1.39° 1.40° 1.41° 1.42° 1.43° 1.44° 1.45° 1.46° 1.47° 1.48° 1.49° 1.50° log sin difi 8.2419 8.2462 8.2505 8.2547 8.2589 8.2630 8.2672 8.2712 8.2753 8.2793 8.2832 i.2872 (.2911 !.2949 S.29S8 S.3025 (.3063 S.3100 (.3137 (.3174 8.3210 8.3246 8.3282 8.3317 8.3353 8.3388 8.3422 8.3456 8.3491 S.3524 8.3558 8.3591 8.3624 8.3657 8.3689 8.3722 8.3754 8.3786 8.3817 8.3S48 5.3880 S.3911 8.3941 8.3972 8.4002 8.4032 8.4062 8.4091 8.4121 8.4150 8.4179 log cos 43 43 42 42 41 42 40 41 40 39 40 39 38 39 37 38 37 37 37 36 36 36 35 36 35 34 34 35 33 34 33 33 33 32 33 32 32 31 31 32 '31 30 31 30 30 30 29 30 29 29 dig. log tan 8.2419 8.2462 8.2505 8.2548 8.2590 8.2631 8.2672 8.2713 8.2754 8.2794 S.2833 8.2S73 8.2912 S.2950 8.2988 8.3026 8.3064 8.3101 8.3138 8.3175 S.3211 8.3247 8.32S3 8.3318 8.3354 8.3389 8.3423 8.3458 8.3492 8.3525 8.3559 8.3592 8.3625 8.3658 8.3691 8.3723 8.3755 8.3787 8.3818 8.3850 S.38S1 8.3912 8.3943 8.3973 8.4003 8.4033 8.4063 8.4093 8.4122 8.4152 8.4181 log cot com. difi. 43 43 43 42 41 41 41 41 40 39 40 39 38 38 38 38 37 37 37 36 36 36 35 36 35 34 35 34 33 34 33 33 33 33 32 32 32 31 32 31 31 31 30 30 30 30 30 29 30 log cot 1.7581 1.7538 1.7495 1.7452 1.7410 1.7369 1.7328 1.7287 1.7246 1.7206 1.7167 1.7127 1.7088 1.7050 1.7012 1.6974 1.6936 1.6899 1.6862 1.6825 1.6789 1.6753 1.6717 1.66S2 1.6646 1.6611 1.6577 1.6542 1.6508 1.6475 1.6441 1.6408 1.6375 1.6342 1.6309 1.6277 1.6245 1.6213 1.6182 1.6150 1.6119 1.6088 1.6057 1.6027 1.5997 1.5967 1.5937 1.5907 1.5878 1.5848 1.5819 com. , difi. l0 & tan 88° log cos 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9.9999 9 9999 9.9999 9 9999 9.9999 9.9999 9.9999 log sin 89.00° 88.99° 88.98° 88.97° 88.96° 88.95° 88.94° 88.93° 88.92° 88.91° 88.90° 88.89° 88.86° 88.85° 88.84° 88.83° 88.82° 88.81° 88.80° 88.79° 88.78° 88.77° 88.76° 88.75° 8S.74° 88.73° 88.72° 8S.71° 88.70° 88.69° 88.68° 88.67° 88.66° 88.65° 88.64° 88.63° 88.62° 88.61° 88.60° 88.59° 88.58° 88.57° 88.56° 88.55° 88.54° 88.53° 88.52° 88.51° 88.50° Angle Flop. Parts Difference 43 42 4.3 4.2 8.6 8.4 12.9 12.6 17.2 16.8 21.5 21.0 25.8 25.2 30.1 29.4 34.4 33.6 38.7 37.8 41 40 4.1 4.0 8.2 8.0 12.3 12.0 16.4 16.0 20.5 20.0 24.6 24.0 28.7 28.0 32.8 32.0 36.9 36.0 39 38 3.9 3.8 7.8 7.6 11.7 11.4 15.6 15.2 19.5 19.0 23.4 22.8 27.3 26.6 31.2 30.4 35.1 34.2 37 36 3.7 3.6 7.4 7.2 11.1 10.8 14.8 14.4 18.5 iao 22.2 21.6 25.9 25.2 29.6 28.8 33.3 32.4 33 34 3.4 3.3 6.8 6.6 10.2 9.9 13.6 13.2 17.0 16.5 20.4 19.8 23.8 23.1 '27.2 26.4 30.6 29.7 31 30 3.1 30 6.2 6.0 9.3 9.0 12.4 12.0 15.5 15.0 18.6 18.0 21.7 21.0 24.8 24.0 27.9 27.0 35 3 5 7.0 10.5 14.0 17.5 21.0 24.5 28.0 31.5 32 3.2 6.4 9.6 12.8 16.0 19.2 22.4 25.6 29 2.9 5.8 8.7 11.6 14.5 17.4 20.3 23.2 26.1 COSINES, TANGENTS, AND COTANGENTS 23 1° Angle log sin dig. log tan com. difi. log cot log COB Prop. Farts 1.50° 8.4179 29 8.4181 29 1.5819 9.9999 88.50° •U 1.51° 8.4208 8.4210 1.5790 9.9998 88.49° bo 1.52° 8.4237 29 8.4238 28 1.5762 9.9998 88.48° 13 DifierAnnn 1.53° 8.4265 28 28 8.4267 29 28 1.5733 9.9998 88.47° 1.54° 8.4293 8.4295 1.5705 9.9998 88.46° W 1.55° 1.56° 8.4322 8.4349 29 27 28 8.4323 8.4351 28 28 28 1.5677 1.5649 9.9998 9.9998 88.45° 88.44° 1.57° 8.4377 8.4379 1.5621 9.9998 88.43° 1.58° 8.4405 26 8.4406 27 1.5594 9.9998 88.42° 1.59° 1.60° 1.61° 8.4432 27 27 27 8.4434 28 27 27 1.5566 9.9998 88.41° 88.40° 88.39° 1 2 3 29 2.9 5.8 8.7 28 2.8 5.6 8.4 8.4459 8.4461 1.5539 9.9998 8.4486 8.4488 1.5512 9.9998 1.62° 8.4513 27 8.4515 27 1.5485 9.9998 88.38° 4 11.6 11.2 1.63° 8.4540 27 27 8.4542 27 26 1.5458 9.9998 88.37° 5 6 14.5 17.4 14.0 16.8 1.64° 8.4567 8.4568 1.5432 9.9998 88.36° 7 20.3 19.6 22.4 25.2 1.65° 8.4593 26 26 8.4595 27 26 1.5405 9.9998 9.9998 88.35° 88.34° 8 9 26.1 1.66° 8.4619 26 8.4621 26 1.5379 1.67° 8.4645 8.4647 1.5353 9.9998 88.33° 27 26 1.68° 8.4671 26 8.4673 26 1.5327 9.9998 88.32° 1 2 3 4 5 6 2.7 5.4 8.1 10.8 13.5 16.2 2.6 5.2 7.8 10.4 13.0 15.6 1.69° 1.70° 1.71° 8.4697 26 26 25 8.4699 26 26 25 1.5301 9.9998 88.31° 88.30° 88.29° 8.4723 8.4725 1.5275 9.9998 8.4748 8.4750 1.5250 9.9998 1.72° 8.4773 25 8.4775 25 1.5225 9.9998 88.28° 7 8 9 18.9 21.6 24.3 18.2 20.8 23.4 1.73° 1.74° 8.4799 8.4824 26 25 8.4801 8.4826 26 25 1.5199 1.5174 9.9998 9.999S 88.27° 88.26° 1.75° 8.4848 24 8.4851 25 1.5149 9.9998 88.25° 25 24 1.76° 8.4873 25 25 8.4875 24 25 1.5125 9.9998 • 88.24° 1 2 2.5 5.0 2.4 ' 4.8 1.77° 8.4898 8.4900 1.5100 9.9998 88.23° 3 7.5 10.0 12.5 7.2 9 6 1.78° 8.4922 24 8.4924 1.5076 9.9998 88.22° 5 12.0 1.79° 8.4947 25 24 8.4949 25 24 1.5051 9.9998 88.21° 6 7 8 9 15.0 17.6 20.0 22.5 14.4 16.8 19.2 21.6 1.80° 1.81° 8.4971 24 8.4973 24 1.5027 9.9998 88.20° 88.19° 8.4995 8.4997 1.5003 9.9998 1.82° 8.5019 24 8.5021 24 1.4979 9.9998 88.18° 1.83° 8.5043 24 8.5045 24 1.4955 9.9998 88.17° 23 22 23 23 1 2.3 2.2 1.84° 8.5066 8.5068 1.4932 9.9998 88.16° 2 4.6 4.4 1.85° 8.5090 24 8.5092 24 1.4908 9.9998 88.15° 3 4 5 6.9 9.2 11.5 6.6 1.86° 8.5113 23 8.5115 23 1.4885 9.9998 88.14° 11.0 1.87° 8.5136 23 8.5139 1.4861 9.9998 88.13° 6 7 13.8 16.1 13.2 15.4 1.88° 8.5160 24 8.5162 23 1.4838 9.9998 88.12° 8 18.4 17.6 1.89° 1.90° 1.91° 8.5183 23 23 22 8.5185 8.5208 23 23 23 1.4815 9.9998 88.11° 88.10° 88.09° 9 20.7 19.8 1 21 2.1 8.5206 1.4792 9.9998 8.5228 8.5231 1.4769 9.9998 1.92° 8.5251 23 8.5253 22 1.4747 9.9998 88.08° 2 4.2 1.93° 8.5274 23 22 8.5276 23 22 1.4724 9.9998 88.07° 3 4 6.3 8.4 1.94° 8.5296 8.5298 1.4702 9.9998 88.06° 5 Q 10.5 12 6 1.95° 8.5318 22 8.5321 23 1.4679 9.9997 88.05° 1 14.7 1.96° 8.5340 22 23 8.5343 22 22 1.4657 9.9997 88.04° 8 9 16.8 18.9 1.97° 8.5363 8.5365 1.4635 9.9997 88.03° 1.98° 8.5385 22 8.5387 22 1.4613 9.9997 88.02° 1.99° 2.00° 8.5406 21 22 8.5409 22 22 1.4591 9.9997 88.01° 88.00° 8.5428 8.5431 1.4569 9.9997 log cos diS. log cot com. difi. log tan log sin Angle 88° 24 TABLE III. LOGARITHMIC SINES 2° Angle log sin diff. log tan com. diff. log cot log cos Proj . Farts 2.00° 2.01° 8.5428 22 8.5431 22 1.4569 9.9997 88.00° 87.99° to © S.5450 8.5453 1.4547 9.9997 2.02° S.5471 SI 8.5474 21 1.4526 9.9997 87.98° Td 2.03° 8.5493 22 21 8.5496 22 21 1.4504 9.9997 87.97° u 2.0V 5 8.5514 8.5517 1.4483 9.9997 87.96° w 1=1 2.0.S° 2.06° 8.5535 8.5557 21 22 21 8.5538 8.5559 21 21 21 1.4462 1.4441 9.9997 9.9997 87.95° 87.94° 22 2.07° 8.5578 8.5580 1.4420 9.9997 87.93° 1 2 2.2 4 4 i 2.08° 8.5598 20 8.5601 21 1.4399 9.9997 87.92° 3 6.8 i 2.09° 8.5619 21 8.5622 21 1.4378 9.9997 87.91° 4 8.8 2.10° 2.11° 21 21 21 21 20 21 87.90° 87.89° 5 6 7 g 11.0 13.2 > 15.4 17 6 i 8.5640 8.5643 1.4357 9.9997 8.5661 8.5664 1.4336 9.9997 2.12° 8.5681 21 8.5684 1.4316 9.9997 87.88° 9 19.8 2.13° 8.5702 20 8.5705 20 1.4295 9.9997 87.87° 21 2 1 r 2.14° 8.5722 8.5725 1.4275 9.9997 87.86° 1 2.15° 8.5742 20 8.5745 20 1.4255 9.9997 87.85° 2 4.2 2.16° 8.5762 20 20 8.5765 20 1.4235 9.9997 87.84° 3 4 5 6.3 8.4 10.5 2.17° 8.5782 8.5785 1.4215 9.9997 87.83° 2.18° 8.5802 20 8.5805 20 1.4195 9.9997 87.82° 6 7 12.6 14 7 2.19° 2.20° 2.21° 8.5822 20 20 20 8.5825 20 20 20 1.4175 9.9997 87.81° 87.80° 87.79° 8 9 16.8 18.9 8.5842 8.5845 8.5865 1.4155 9.99-97 20 8.5862 1.4135 9.9997 2.22° 8.5881 19 8.5884 19 1.4116 9.9997 87.78° 1 2.0 2.23° 8.5901 20 19 8.5904 20 19 1.4096 9.9997 87.77° 2 3 4.0 ( 6.0 2.24° 8.5920 8.5923 1.4077 9.9997 87.76° 4 8.0 2.25° 8.5939 19 8.5943 20 1.4057 9.9997 87.75° 5 Q 10.0 2.26° 8.5959 20 19 8.5962 19 • 19 1.4038 9.9997 87.74° 7 8 14.0 16.0 2.27° 8.5978 19 8.5981 19 1.4019 9.9997 87.73° 9 18.0 2.28° 8.5997 8.6000 1.4000 9.9997 87.72° 2.29° 2.30° 2.31° 8.6016 19 19 19 8.6019 19 19 19 1.3981 9.9997 87.71° 87.70° 87.69° 1 2 3 19 1.9 3.8 5.7 8.6035 8.6038 1.3962 9.9996 8.6054 8.6057 1.3943 9.9996 2.32° 8.6072 18 8.6076 19 1.3924 9.9996 87.68° 4 5 7.6 9.B !; 11.4 13.3 15.2 2.33° 2.34° 8.6091 8.6110 19 19 ' 8.6095 8.6113 19 18 1.3905 1.3887 9.9996 9.9996 87.67° 87.66° 6 7 8 2.35° 2.36° 8.6128 8.6147 18 19 18 8.6132 8.6150 19 18 19 1.3868 1.3850 9.9996 9.9996 87.65° 87.64° 9 17.1 18 1.8 2.37° 8.6165 8.6169 1.3831 9.9996 87.63° 1 2.38° 8.6183 18 8.6187 18 1.3813 9.9996 87.62° 2 3.6 2.39° 2.40° 8.6201 18 19 8.6205 18 18 1.3795 9.9996 87.61° 87.60° 3 4 5 5.4 7.2 9.0 8.6220 8.6223 1.3777 9.9996 18 19 6 7 10.8 12.6 2.41° 8.6238 8.6242 1.3758 9.9996 87.59° 2.42° 8.6256 18 8.6260 1.3740 9.9996 87.58° 8 14.4 2.43° 8.6274 18 8.6277 17 1.3723 9.9996 87.57° 9 16.2 2.44° 8.6291 17 8.6295 18 1.3705 9.9996 87.56° 17 2.45° 8.6309 18 8.6313 18 1.3687 9.9996 87.55° 1 1.7 2.46° 8.6327 18 17 8.6331 18 17 1.3669 9.9996 87.54° 2 3 3.4 5.1 2.47° 8.6344 8.6348 1.3652 9.9996 87.53° 4 5 6 6.8 8.5 10.2 2.48° 8.6362 18 8.6366 18 1.3634 9.9996 87.52° 2.49° 2.50° 8.6379 17 18 8.6384 18 17 1.3616 9.9996 87.51° 87.50° 7 8 9 11.9 13.6 15.3 i 8.6397 8.6401 1.3599 9.9996 log cos diff. log cot com. diff. log tan log sin Angle 87° COSINES, TANGENTS, AND COTANGENTS 25 2° Angle log sin cliff. log tan com. diS. log cot log cos Prop. Faits 2.50° 2.51° 2.52° 2.53° 2.54° 2.55° 2.56° 2.57° 2.58° 2.59° 2.60° 2.61° 2.62° 2.63° 2.64° 2.65° 2.66° 2.67° 2.68° 2.69° 2.70° 2.71° 2.72° 2.73° 2.74° 2.75° 2.76° 2.77° 2.78° 2.79° 2.80° 2.81° 2.82° 2.83° 2.84° 2.85° 2.86° 2.87° 2.88° 2.89° 2.90° 2.91° 2.92° 2.93° 2.94° 2.95° 2.96° 2.97° 2.98° 2.99° 3.00° 8.6397 17 17 18 17 17 17 17 17 16 17 17 16 17 16 17 16 16 17 16 16 16 16 16 16 15 16 16 16 15 16 15 16 15 15 15 16 15 15 15 15 15 15 15 14 15 15 14 15 15 14 8.6401 17 18 17 17 17 17 17 17 17 16 17 17 16 17 16 17 16 16 16 17 16 16 16 16 15 16 16 16 15 16 15 16 15 16 15 15 15 15 15 15 15 15 15 15 15 15 14 15 14 15 1.3599 9.9996 87.50° 87.49° 87.48° 87.47° 87.46° 87.45° 87.44° 87!43° 87.42° 87.41° 87.40° 87.39° 87.38° 87.37° 87.36° 87.35° 87.34° 87.33° 87.32° 87.31° 87.30° 87.29° 87.28° 87.27° 87.26° 87.25° 87.24° 87.23° 87.22° 87.21° 87.20° 87.19° 87.18° 87.17° 87.16° 87.15° 87.14° 87.13° 87.12° 87.11° 87.10° 87.09° 87.08° 87.07° 87.06° 87.05° 87.04° 87.03° 87.02° 87.01° 87.00° £ 60 •a a m W B a n iH •r-t 8.6414 8.6431 8.6449 8.6466 8.6483 8.6500 8.6517 8.6534 8.6550 8.6418 8.6436 8.6453 8.6470 8.6487 8.6504 8.6521 8.6538 8.6555 1.3582 1.3564 1.3547 1.3530 1.3513 1.3496 1.3479 1.3462 1.3445 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 9.9996 i 2 3 4 5 6 7 8 9 18 1.8 3.6 5.4 7.2 9.0 10.8 12.6 14.4 16.2 i 8.6567 8.6571 1.3429 9.9996 8.6584 8.6600 8.6617 8.6633 8.6650 8.6666 8.6682 8.6699 8.6715 8.6588 8.6605 8.6621 8.6638 8.6654 8.6671 8.6687 8.6703 8.6719 1.3412 1.3395 1.3379 1.3362 1.3346 1.3329 1.3313 1.3297 1.3281 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 1 2 3 4 5 6 7 8 9 17 1.7 3.4 5.1 6.8 8.5 10.2 11.9 13.6 15.3 8.6731 8.6736 1.3264 9.9995 8.6747 8.6763 8.6779 8.6795 8.6810 8.6826 8.6842 8.6858 8.6873 8.6752 8.6768 8.6784 8.6800 8.6815 8.6831 8.6847 8.6863 8.6878 1.3248 1.3232 1.3216 1.3200 1.3185 1.3169 1.3153 1.3137 1.3122 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 1 2 3 4 5 6 7 8 9 16 1.6 3.2 4.8 6.4 8.0 9.6 11.2 12.8 14.4 8.6889 8.6894 1.3106 9.9995 8.6904 8.6920 8.6935 8.6950 8.6965 8.6981 8.6996 8.7011 8.7026 8.6909 8.6925 8.6940 8.6956 8.6971 8.6986 8.7001 8.7016 8.7031 1.3091 1.3075 1.3060 1.3044 1.3029 1.3014 1.2999 1.2984 1.2969 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9995 9.9994 1 2 3 4 5 6 7 8 9 15 1.6 3.0 4.5 6.0 7.6 9.0 10.6 12.0 13.5 8.7041 8.7046 1.2954 9.9994 1 2 3 4 5 6 7 8 9 14 1.4 2.8 4.2 6.6 7.0 8.4 9.8 11.2 12.6 8.7056 8.7071 8.7086 8.7100 8.7115 8.7130 8.7144 8.7159 8.7174 8.7061 8.7076 8.7091 8.7106 8.7121 8.7136 8.7150 8.7165 8.7179 1.2939 1.2924 1.2909 1.2894 1.2879 1.2864 1.2850 1.2835 1.2821 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 8.7188 8.7194 1.2806 9.9994 log cos diS. log cot com. diS. log tan log sin Angle 87° 26 TABLE III. LOGARITHMIC SINES 3° Angle log sin difi. log tan com. difi. log cot log cos Prop. Farts 3.00° 3.01° 3.02° 3.03° 3.04° 3.05° 3.06° 8.7188 14 15 14 14 15 14 14 8.7194 14 15 14 15 14 14 14 1.2806 9.9994 87.00° 86.99° 86.98° 86.97° 86.96° 86.95° 86.94° ■ »H bo ■a 03 u m o a at s m P 8.7202 8.7217 8.7231 8.7245 8.7260 8.7274 8.7208 8.7223 8.7237 8.7252 8.7266 8.7280 1.2792 1.2777 1.2763 1.2748 1.2734 1.2720 9.9994 9.9994 9.9994 9.9994 9.9994 9.9994 3.07° 3.08° 3.09° 3.10° 3.11° 3.12° 3.13° 8.7288 8.7302 8.7316 14 ' 14 14 14 14 14 14 8.7294 8.7308 8.7323 14 15 14 14 14 14 13 1.2706 1.2692 1.2677 9.9994 9.9994 9.9994 86.93° 86.92° 86.91° 86.90° 86.89° 86.88° 86.87° 1 2 3 4 5 6 7 8 9 1.5 3.0 4.5 6.0 7.5 9.0 10.5 12.0 13.5 8.7330 8.7337 1.2663 9.9994 8.7344 8.7358 8.7372 8.7351 8.7365 8.7379 1.2649 1.2635 1.2621 9.9994 9.9994 9.9994 3.14° 3.15° 3.16° 8.7386 8.7400 8.7413 14 13 14 8.7392 8.7406 8.7420 14 14 14 1.2608 1.2594 1.2580 9.9993 9.9993 9.9993 86.86° S6.85° 86.84° 3.17° 3.18° 3.19° 3.20° 3.21" 3.22° 3.23° 8.7427 8.7441 8.7454 14 13 14 14 13 13 ■ 14 8.7434 8.7448 8.7461 14 13 14 13 14 13 14 1.2566 1.2552 1.2539 9.9993 9.9993 9.9993 86.83° 86.82° 86.81° 86.80° 86.79° 86.78° S6.77° 1 2 3 4 5 6 7 8 9 14 1.4 2.8 4.2 5.6 7.0 8.4 9.8 11.2 12.6 8.7468 8.7475 1.2525 9.9993 8.7482 8.7495 8.7508 8.7488 8.7502 8.7515 1.2512 1.2498 1.2485 9.9993 9.9993 9.9993 3.24° 3.25° 3.26° 8.7522 8.7535 8.7549 13 14 13 8.7529 8.7542 8.7556 13 14 13 1.2471 1.2458 1.2444 9.9993 9.9993 9.9993 86.76° 86.75° 86.74° 3.27° 3.28° 3.29° 3.30° 3.31° 3.32° 3.33° 3.34° 3.35° 3.36° 8.7562 8.7575 8.7588 8.7602 8.7615 8.7628 8.7641 8.7654 8.7667 8.7680 13 13 14 13 13 13 13 13 13 13 8.7569 8.7582 8.7596 13 14 13 13 13 13 13 13 13 13 1.2431 1.2418 1.2404 9.9993 9.9993 9.9993 86.73° 86.72° 86.71° 86.70° 86.69° 86.68° 86.67° 86.66° 86.65° 86.64° 1 2 3 4 5 6 7 8 9 ~13" 1.3 2.6 3.9 5.2 6.5 7.8 9.1 10.4 11.7 8.7609 1.2391 9.9993 8.7622 8.7635 8.7648 8.7661 8.7674 8.7687 1.2378 1.2365 1.2352 1.2339 1.2326 1.2313 9.9993 9.9993 9.9993 9.9993 9.9993 9.9993 3.37° 3.38° 3.39° 3.40° 3.41° 3.42° 3.43° 8.7693 8.7705 8.7718 12 13 13 13 12 13 13 8.7700 8.7713 8.7726 13 13 13 12 13 13 13 1.2300 1.22S7 1.2274 9.9992 9.9992 9.9992 86.63° 86.62° 86.61° 86.60° 86.59° 86.58° 86.57° 1 2 3 4 5 6 ~12~ 1.2 2.4 3.6 4.8 6.0 7.2 8.7731 8.7739 1.2261 9.9992 8.7744 8.7756 8.7769 8.7751 8.7764 8.7777 1.2249 1.2236 1.2223 9.9992 9.9992 9.9992 3.44° 3.45° 3.46° 8.7782 8.7794 8.7807 12 13 12 8.7790 8.7802 8.7815 12 13 12 1.2210 1.2198 1.2185 9.9992 9.9992 9.9992 86.56° 86.55° 86.54° 7 8 9 8.4 9.6 10.8 3.47° 3.48° 3.49° 3.50° 8.7819 8.7832 8.7844 13 12 13 8.7827 8.7840 8.7852 13 12 13 1.2173 1.2160 1.2148 9.9992 9.9992 9.9992 86.53° 86.52° 86.51° 86.50° 8.7857 8.7865 1.2135 9.9992 log COB difi. log cot com. difi. log tan log sin Angle 86° COSINES, TANGENTS, AND COTANGENTS Angle 3.50° 3.51° 3.52° 1.53° 3.54° 3.55° 3.56° 3.57° 5.58° 3.59° 3.60° 3.61° 5.62° 3.63° 3.64° 3.65° 3.66° 3.67° 3.6S° 3.69° 3.70° o.71° 3.72° 3.73° 3.74° 3.75° 3.76° 3.77° 3.7S° 3.79° 3.80° 3.81° 3.S2° 3.S3° 3.S4° 3.85° 3.36° 3.S7° 3.SS° 3.S9° 3.90° 3.91° 3.92° 3.93° 3.94° 3.95° 3.96° 3.97° 3.98° 3.99° 4.00° log sin difi. S.7S57 S.7S69 S.7SS1 S.7S94 S.7906 S.791S &7930 S.7943 S.7955 S.7967 S.7979 I S.7991 S.S003 S.S015 S.S027 S.S039 S.S051 S.S062 S.S074 S.80S6 S.S09S S.S109 S.S121 S.S133 S.S144 S.S156 S.S168 S.S179 S.S191 S.S202 S.S213 S.S22. S.8236 S.S24S S.S259 S.S270 S.S2S1 S.S293 aS304| S.S315 S.8526 I S.S34S S.S359 S.S370 S.S3S1 S.S392 S.S403 I S.S414 | S.S425 S.S436 13 13 13 13 19 13 13 13 13 13 13 13 13 13 ia 13 11 13 13 19 11 13 19 11 13 13 11 19 11 11 13 11 13 11 11 11 19 11 11 11 11 11 11 11 11 11 11 11 11 11 log cos i difi. log tan ! com. difi. i log cot S.7S65 8.7S77 S.7S90 a?902 S.7914 S.7927 , S.7939 | S.7951 S.7963 S.7975 S.798S s.sooo| S.S012 ' S.S024 S.S036 I S.S04S I S.S059 S.S071 S S.S0S3 S.S095 S.S107 8.8119 | S.S130 ! S.S142 S.S154 S.S165 S.S177 S.S1SS S.S200 S.S212 S.S223 ! S.S234 S.S246 S.S257 S.S269 S.S2S0 S.S291 S.S302 | S.S314 S.S325 S.S336 S.S347 S.S35S S.S370 | S.S3S1 S.S392 S.8403 S.S4H! S.S425 ' S.8436 S.S446 19 13 13 13 13 13 19 13 13 13 19 19 19 13 13 11 13 13 13 13 19 11 13 19 II 19 II 13 19 11 II 19 11 13 11 11 11 13 11 11 11 11 13 11 11 11 11 II 11 10 1.2135 1.2123 1.2110 1.209S 1.2086 1.2073 1.2061 1.2049 1.2037 1.2025 1.2012 1.2000 1.19S8 1.1976 1.1964 1.1952 1.1941 1.1929 1.1917 1.1905 1.1S93 1.1SS1 1.1S70 1.1S5S 1.1S46 1.1S35 1.1 S23 1.1S12 1.1S00 1.1 7SS 1.17 1.1766 1.1754 1.1743 I 1.1731 1.1720 J 1.1709 : 1.169S I 1.16S6 1.1675 1.1664 I 1.1653 i 1.1642 [ 1.1630 ! 1.1619 1.160S 1.1597 | 1.15S6 i 1.1575 1.1564 1.1554 log cot com. difi. log cos 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9992 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9991 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.9990 9.99S9 9.99S9 log tan log sin 86° 86.50° S6.49° S6.4S° S6.47° 86.46° 86.45° 86.44° S6.43° S6.42 S6.41° 86.40° S6.39° S6.3S° S6.37° S6.36° S6.35° 86.34° S6.33° S6.32° S6.31° 86.30° 86.29° S6.2S° S6.27° 86.26° 86.25° S6.24° S6.23° S6.22° S6.21° 86.20° 86.19° S6.1S° 86.17° 86.16° S6.15° S6.14° S6.13° 86.12° S6.11° 86.10° S6.09 S6.0S S6.07 S6.06° S6.05° S6.04° S6.03° S6.02° S6.01° 86.00° Angle Prop. Farts T* - 13 i-i 2.6 3.9 5.2 6.5 7.S 9.1 10.4 11.7 12 ! 1.2 2.4 3.6 I 4.S 6.0 i 7.2 I °" 4 9.6 10.S 11 1.1 2.2 as 4.4 5.5 6.6 S.S 9.9 10 1 1.0 '"> 2.0 o 3.0 4 4.0 5 5.0 6 6.0 l 7.0 S 8.0 9 9.0 28 TABLE III. LOGAKITHMIC SINES Angle log sin difi. log tan com. difi. log cot log COB Prop. Parts 4.00° 4.01° 4.02° 4.03° 4.04° 4.05° 4.06° 4.07° 4.08° 4.09° 4.10° 4.11° 4.12° 4.13° 4.14° 4.15° 4.16° 4.17° 4.18° 4.19° 4.20° 4.21° 4.22° 4.23° 4.24° 4.25° 4.26° 4.27° 4.28° 4.29° 4.30° 4.31° 4.32° 4.33° 4.34° 4.35° 4.36° 4.37° 4.38° 4.39° 4.40° 4.41° 4.42° 4.43° 4.44° 4.45° 4.46° 4.47° 4.48° 4.49° 4.50° 8.8436 8.8447 8.8457 8.8468 8.8479 8.8490 8.8500 8.8511 8.8522 8.8532 8.8543 8.8553 8.8564 8.8575 8.8585 8.8595 8.8606 8.8616 8.8627 8.8637 8.8647 8.8658 8.8668 8.8678 8.8688 8.8699 8.8709 8.8719 8.8729 8.8739 8.8749 S.8759 5.8769 5.8780 S.S790 S.8799 S.8809 5.8819 5.8829 5.8839 8.8849 5.8859 5.8869 5.8878 8.8898 8.8908 8.8917 8.8927 8.8937 8.8946 n 10 11 ii n 10 11 11 10 11 10 11 11 10 10 11 10 11 10 10 11 10 10 10 11 10 10 10 10 10 10 10 11 10 9 10 10 10 10 10 10 10 9 10 10 10 9 10 10 9 8.8446 8.8457 8.8468 8.8479 8.8490 8.8501 8.8511 8.8522 8.8533 8.8543 8.8554 5.8565 5.8575 5.8586 5.8596 5.8607 5.8617 5.8628 5.8638 5.8649 8.8659 8.8669 8.8680 8.8690 8.8700 8.8711 8.8721 8.8731 8.8741 8.8751 8.8762 8.8772 8.8782 8.8792 8.8802 8.8812 8.8822 8.8832 8.8842 8.8852 8.8862 8.8872 8.8882 8.8891 8.8901 8.8911 8.8921 8.8931 8.8940 8.8950 8.8960 n n n n n 10 n n 10 n n 10 n 10 11 10 11 10 11 10 10 11 10 10 11 10 10 10 10 11 10 10 10 10 10 10 10 10 10 10 10 10 9 10 10 10 10 9 10 10 1.1554 9.9989 1.1543 1.1532 1.1521 1.1510 1.1499 1.1489 1.1478 1.1467 1.1457 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 1.1446 9.9989 1.1435 1.1425 1.1414 1.1404 1.1393 1.1383 1.1372 1.1362 1.1351 9.9989 9.9989 9.9989 9.9989 9.9989 9.9989 9.9988 9.9988 9.9988 1.1341 9.9988 1.1331 1.1320 1.1310 1.1300 1.1289 1.1279 1.1269 1.1259 1.1249 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 9.9988 1.1238 9.9988 1.122S 1.1218 1.1208 1.1198 1.1188 1.1178 1.1168 1.1158 1.1148 9.9988 9.9988 9.9988 9.9988 9.9987 9.9987 9.9987 9.9987 9.9987 1.1138 9.9987 1.1128 1.1118 1.1109 1.1099 1.1089 1.1079 1.1069 1.1060 1.1050 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 9.9987 1.1040 9.9987 86.00° 85.99° 85.98° 85.97° 85.96° 85.95° 85.94° 85.93° 85.92° 85.91° 85.90° 85.89° 85.88° 85.87° 85.86° 85.85° 85.84° 85.83° 85.82° 85.81° 85.80° 85.79° 85.78° 85.77° 85.76° 85.75° 85.74° 85.73° 85.72° 85.71° 85.70° 85.69° 85.68° 85.67° 85.66° 85.65° 85.64° 85.63° 85.62° 85.61° 85.60° 85.59° 85.58° 85.57° 85.56° 85.55° 85.54° 85.53° 85.52° 85.51° 85.50° log cos difi. log cot Sj?' log tan log sin Angle 85 p be ■■a a M w 11 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 10 1 1.0 2 2.0 3 3.0 4 4.0 6 6.0 6 6.0 7 7.0 8 8.0 9 9.0 9 1 0.9 2 1.8 3 2.7 4 3.6 5 4.5 6 5.4 7 6.3 8 7.2 9 8.1 COSINES, TANGENTS, AND COTANGENTS 29 Angle 4.50° 4.51° 4.52° 4.53° 4.54° 4.55° 4.56° 4.57° 4.58° 4.59° 4.60° 4.61° 4.62° 4.63° 4.64° 4.65° 4.66° 4.67° 4.68° 4.69° 4.70° 4.71° 4.72° 4.73° 4.74° 4.75° 4.76° 4.77° 4.78° 4.79° 4.80° 4.81° 4.82° 4.83° 4.84° 4.85° 4.86° 4.87° 4.8S° 4.89° 4.90° 4.91° 4.92° 4.93° 4.94° 4.95° 4.96° 4.97° 4.98° 4.99° 5.00° log sin 8.8946 3.8956 i.8966 5.8975 i.8985 S.8994 5.9004 S.9013 S.9023 S.9032 8.9042 diff. 8.9051 8.9060 8.9070 8.9079 8.9089 8.9098 8.9107 8.9116 8.9126 8.9135 8.9144 8.9153 8.9162 8.9172 8.9181 8.9190 8.9199 8.9208 8.9217 8.9226 8.9235 8.9244 8.9253 8.9262 8.9271 8.9280 8.9289 8.9298 8.9307 8.9315 8.9324 8.9333 8.9342 8.9351 8.9359 8.9368 8.9377 8.9386 8.9394 8.9403 10 10 9 10 9 10 9 10 9 10 10 9 10 10 9 9 9 9 10 log tan 8.8960 8.8970 8.8979 8.8989 8.8998 8.9008 8.9018 8.9027 8.9037 8.9046 8.9056 8.9065 8.9075 8.9084 8.9093 8.9103 8.9112 8.9122 8.9131 8.9140 8.9150 8.9159 8.9168 8.9177 8.9186 8.9196 8.9205 8.9214 8.9223 8.9232 8.9241 log cos diS. log cot 8.9250 8.9260 8.9269 8.9278 8.9287 8.9296 8.9305 8.9313 8.9322 8.9331 8.9340 8.9349 8.9358 8.9367 8.9376 8.9384 8.9393 8.9402 8.9411 8.9420 com. difi. 10 9 10 9 10 10 9 10 9 10 10 9 10 9 9 10 9 9 10 9 9 9 9 9 9 10 9 9 dm. log cot 1.1040 1.1030 1.1021 1.1011 log cos 9.9987 9.9987 9.9986 9.9986 1.1002 9.9986 1.0992 9.9986 1.0982 9.9986 1.0973 1.0963 1.0954 1.0944 1.0935 1.0925 1.0916 1.0907 1.0897 1 1.0878 1.0869 1.0860 1.0850 1.0841 1.0832 1.0823 1.0814 1.0804 1.0795 1.0786 1.0777 1.0768 1.0759 1.0750 1.0740 1.0731 1.0722 1.0713 1.0704 1.0695 1.0687 1.0678 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9986 9.9985 9.9985 9.9985 9.9985 9.9985 9.99S5 9.9985 9.9985 9.9985 9.9985 9.9985 9.9985 9.9985 1.0669 1.0660 1.0651 1.0642 1.0633 1.0624 1.0616 1.0607 1.0598 1.0589 9.9985 9.9985 9.9985 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 9.9984 1.0580 log tan 85° 9.9983 log sin 85.50° 85.49° 85.48° 85.47° 85.46° 85.45° 85.44° 85.43° 85.42° 85.41° 85.40° 85.39° 85.38° 85.37° 85.36° 85.35° 85.34° 85.33° 85.32° 85.31° 85.30° 85.29° 85.28° 85.27° 85.26° 85.25° 85.24° 85.23° 85.22° 85.21° 85.20° 85.19° 85.18° 85.17° 85.16° 85.15° 85.14° 85.13° 85.12° 85.11° 85.10° 85.09° 85.08° 85.07° 85.06° 85.05° 85.04° 85.03° 85.02° 85.01° 85.00° Angle Prop. Farts bo & sa 10 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 9 1 0.9 2 1.8 3 2.7 4 3.6 5 4.5 6 5.4 7 6.3 8 7.2 9 8.1 8 1 0.8 2 1.6 3 2.4 4 3.2 5 4.0 6 4.8 7 6.6 8 6.4 9 7.2 30 TABLE III. LOGARITHMIC SINES 5°-10° Angle log sin dig. log tan com. difi. log cot log cos difi. Prop. Parts 5.0° 5.1° S.9403 86 8.9420 86 1.0580 9.9983 85.0° 84.9° £ ■ i-t 8.9489 8.9506 1.0494 9.9983 5.2° 8.9573 84 8.9591 85 1.0409 9.9982 84.8° T* Difference 5.3° 8.9655 81 8.9674 83 82 1.0326 9.9981 84.7° a 5.4° 8.9736 8.9756 1.0244 9.9981 84.6° w 5 5° 8.9816 80 8.9836 80 1.0164 9.9980 84.5° 5.6° 8.9894 78 76 8.9915 79 77 1.0085 9.9979 84.4° 62 61 60 5.7° 8.9970 8.9992 1.0008 9.9978 84.3° i 6.2 12.4 6.1 12.2 6.0 12.0 5.8° 9.0046 76 9.0068 76 0.9932 9.9978 84.2° 3 18.6 18.3 18.0 5.9° 9.0120 74 72 9.0143 75 73 0.9857 9.9977 84.1° 4 24.8 31.0 24.4 30.5 24.0 30.0 6.0° 9.0192 72 9.0216 73 0.9784 9.9976 84.0° 6 37.2 36.6 36.0 6.1° 9.0264 9.0289 0.9711 9.9975 83.9° 7 8 43.4 49.6 42.7 48.8 42.0 48.0 6.2° 6.3° 9.0334 9.0403 69 9.0360 9.0430 70 0.9640 0.9570 9.9975 9.9974 83.8° 83.7° 9 55.8 54.9 54.0 69 69 W 58 57 6.4° 9.0472 9.0499 0.9501 9.9973 83.6° 1 5.9 5.8 5.7 6.5° 9.0539 67 9.0567 0.9433 9.9972 83.5° 2 11.8 11.6 11.4 6.6° 9.0605 66 65 9.0633 66 66 0.9367 9.9971 83.4° 3 4 17.7 23.6 17.4 23.2 17.1 22.8 6.7° 9.0670 9.0699 0.9301 9.9970 83.3° 5 29.5 29.0 28.5 6.8° 9.0734 64 9.0764 65 0.9236 9.9969 83.2° 6 7 35.4 41.3 34.8 40.6 34.2 39.9 6.9° 9.0797 63 62 9.0828 63 0.9172 9.9968 83.1° 8 47.2 46.4 45.6 7.0° 7.1° 9.0859 61 9.0891 63 0.9109 9.9968 83.0° 82.9° 9 53.1 51.3 56 55 54 9.0920 9.0954 0.9046 9.9967 7.2° 9.0981 61 9.1015 61 0.8985 9.9966 82.8° 1 5.6 5.5 5.4 7.3° 9.1040 59 59 9.1076 61 59 0.8924 9.9965 82.7° 2 3 11.2 16.8 11.0 16.5 10.8 16.2 7.4° 9.1099 9.1135 0.8865 9.9964 82.6° 4 22.4 22.0 21.6 7.5° 9.1157 58 9.1194 59 0.8806 9.9963 82.5° 5 6 28.0 33.6 27.5 33.0 27.0 32.4 7.6° 9.1214 57 57 9.1252 58 0.8748 9.9962 82.4° 7 8 39.2 448 38.5 440 37.8 43.2 7.7° 7.8° 9.1271 9.1326 55 9.1310 9.1367 57 0.8690 0.8633 9.9961 9.9960 82.3° 82.2° 9 50.4 49.5 48.6 7.9° 9.1381 55 55 9.1423 56 55 0.8577 9.9959 82.1° 53 52 51 8.0° 8.1° 9.1436 53 9.1478 55 0.8522 9.9958 82.0° 81.9° 1 2 3 5.3 10.6 15.9 5.2 10.4 15.6 5.1 10.2 15.3 9.1489 9.1533 0.8467 9.9956 8.2° 9.1542 53 9.1587 54 0.8413 9.9955 81.8° 4 1 21.2 26.5 20.8 26.0 20.4 25.5 8.3° 9.1594 52 52 9.1640 53 0.8360 9.9954 81.7° 6 7 31.8 37,1 31.2 36.4 30.6 35.7 8.4° 9.1646 9.1693 0.8307 9.9953 81.6° 8 42.4 41.6 40.8 8.5° 9.1697 51 9.1745 52 0.8255 9.9952 81.5° 9 47.7 46.8 45.9 8.6° 9.1747 50 50 9.1797 52 51 0.8203 9.9951 81.4° 50 49 48 8.7° 9.1797 9.1848 0.8152 9.9950 81.3° 1 5.0 4.9 4.8 8.8° 9.1847 50 9.1898 50 0.8102 9.9949 81.2° 2 10.0 9.8 9.6 8.9° 9.0° 9.1° 9.1895 48 48 48 9.1948 50 49 49 0.8052 9.9947 2 81.1° 81.0° 80.9° 3 4 5 6 7 15.0 20.0 25.0 30.0 35.0 14.7 19.6 24.5 29.4 34.3 14.4 19.2 24.0 28.8 33.6 9.1943 9.1997 0.8003 9.9946 9.1991 9.2046 0.7954 9.9945 9.2° 9.2038 47 9.2094 48 0.7906 9.9944 80.8° 8 40.0 39.2 38.4 9 3° 9.2085 9.2131 47 9.2142 9.2189 48 0.7858 0.7811 9.9943 9.9941 80.7° 80.6° 9 45.0 44.1 43.2 9.4° 46 47 2 47 46 45 9.5° 9.2176 45 9.2236 47 0.7764 9.9940 80.5° 1 4.7 4.6 4.5 9.6° 9.2221 45 45 9.2282 46 46 0.7718 9.9939 2 80.4° 2 3 9.4 14.1 9.2 13.8 9.0 13.5 9.7° 9.2266 9.2328 0.7672 9.9937 80.3° 4 5 18.8 23.5 18.4 23.0 18.0 22.5 9.8° 9.2310 9.2374 46 0.7626 9.9936 80.2° 6 28.2 27.6 27.0 9.9° 10.0° 9.2353 9.2397 43 44 9.2419 45 44 0.7581 9.9935 S0.1° 80.0° 7 8 9 32.9 37.6 42.3 32.2 36.8 41.4 31.5 36.0 40.5 9.2463 0.7537 9.9934 log cos difi. log cot com. din". log tan log sin difi. Angle 80°- 85° COSINES, TANGENTS, AND COTANGENTS 31 10°-15° Angle 10.0° 10.1° 10.2° 10.3° 10.4° 10.5° 10.6° 10.7° 10.8° 10.9° 11.0° 11.1° 11.2° 11.3° 11.4° 11.5° 11.6° 11.7° 11.8° 11.9° 12.0° 12.1° 12.2° 12.3° 12.4° 12.5° 12.6° 12.7° 12.8° 12.9° 13.0° 13.1° 13.2° 13.3° 13.4° 13.5° 13.6° 13.7° 13.8° 13.9° 14.0° 14.1° 14.2° 14.3° 14.4° 14.5° 14.6° 14.7° 14.8° 14.9° 15.0° log sin 9.2397 9.2439 9.2482 9.2524 9.2565 9.2606 9.2647 9.2687 9.2727 9.2767 9.2806 9.2845 9.2883 9.2921 9.2959 9.2997 9.3034 9.3070 9.3107 9.3143 9.3179 9.3214 9.3250 9.3284 9.3319 9.3353 9.3387 9.3421 9.3455 9.3488 9.3521 9.3554 9.3586 9.3618 9.3650 9.3682 9.3713 9.3745 9.3775 9.3806 9.3837 9.3867 9.3897 9.3927 9.3957 9.3986 9.4015 9.4044 9.4073 9.4102 9.4130 diH. log tan com. diS. 9.2463 9.2507 9.2551 9.2594 9.2637 9.2680 2.2722 9.2764 9.2S05 9.2846 9.2887 9.2927 9.2967 9.3006 9.3046 9.3085 9.3123 9.3162 9.3200 9.3237 9.3275 9.3312 9.3349 9.3385 9.3422 9.3458 9.3493 9.3529 9.3564 9.3599 9.3634 9.3668 9.3702 9.3736 9.3770 9.3804 9.3837 9.3870 9.3903 9.3935 9.3968 9.4000 9.4032 9.4064 9.4095 9.4127 9.4158 9.4189 9.4220 9.4250 9.4281 log cot 0.7537 0.7493 0.7449 0.7406 0.7363 0.7320 0.7278 0.7236 0.7195 0.7154 0.7113 0.7073 0.7033 0.6994 0.6954 0.6915 0.6877 0.6838 0.6800 0.6763 0.6725 0.6688 0.6651 0.6615 0.6578 0.6542 0.6507 0.6471 0.6436 0.6401 0.6366 0.6332 0.6298 0.6264 0.6230 0.6196 0.6163 0.6130 0.6097 0.6065 0.6032 0.6000 0.5968 0.5936 0.5905 0.5873 0.5842 0.5811 0.5780 0.5750 0.5719 log COB 9.9934 9.9932 9.9931 9.9929 9.9928 9.9927 9.9925 9.9924 9.9922 9.9921 9.9919 9.9918 9.9916 9.9915 9.9913 9.9912 9.9910 9.9909 9.9907 9.9906 9.9904 9.9902 9.9901 9.9899 9.9897 9.9896 9.9894 9.9892 9.9S91 9.9S89 9.9887 9.9885 9.9884 9.9882 9.9880 9.9878 9.9876 9.9875 9.9873 9.9871 9.9869 9.9867 9.9865 9.9863 9.9861 9.9859 9.9857 9.9855 9.9853 9.9851 9.9849 diH 80.0° 79.9° 79.8° 79.7° 79.6° 79.5° 79.4° 79.3° 79.2° 79.1° 79.0° 78.9° 78.8° 78.7° 78.6° 78.5° 78.4° 78.3° 78.2° 78.1° 78.0° 77.9° 77.8° 77.7° 77.6° 77.5° 77.4° 77.3° 77.2° 77.1° 77.0° 76.9° 76.8° 76.7° 76.6° 76.5° 76.4° 76.3° 76.2° 76.1° 76.0° 75.9° 75.8° 75.7° 75.6° 75.5° 75.4° 75.3° 75.2° 75.1° 75.0° Prop. Parts Difference 44 43 4.4 4.3 8.8 8.6 13.2 12.9 17.6 17.2 22.0 21.5 26.4 25.8 30.8 30.1 35.2 34.5 39.6 38.8 41 40 4.1 4.0 8.2 8.0 12.3 12.0 16.4 16.0 20.5 20.0 24.6 24.0 28.7 28.0 32.8 32.0 36.9 36.0 38 37 3.8 3.7 7.6 7.4 11.4 11.1 15.2 14.8 19.0 18.5 22.8 22.2 26.6 25.9 30.4 29.6 34.2 33.3 35 34 3.5 3.4 7.0 6.8 10.5 10.2 14.0 13.6 17.5 17.0 21.0 20.4 24.5 23.8 28.0 27.2 31.5 30.6 32 31 3.2 3.1 6.4 6.2 9.0 9.3 12.8 12.4 16.0 15.5 19.2 18.6 22.4 21.7 25.6 24.8 28.8 27.9 29 28 2.9 2.8 5.8 5.6 8.7 8.4 11.6 11.2 14.5 14.0 17.4 16.8 20.3 19.6 23.2 22.4 26.1 25.2 42 4.2 8.4 12.6 16.S 21.0 25.2 29.4 33.6 37.8 39 3.9 7.8 11.7 15.6 19.5 23.4 27.3 31.2 35.1 36 3.6 7.2 10.8 14.4 18.0 21.6 25.2 28.8 32.4 33 3.3 6.6 9.9 13.2 16.5 19.8 23.1 26.4 29.7 30 3.0 6.0 9.0 12.0 15.0 18.0 21.0 24.0 27.0 2 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.* log cos difi. log cot ,^ m- log tan log sin diH. Angle 75°-80° 32 TABLE III. LOGARITHMIC SINES 15°- 20" Angle log sin diS. log tan com, diff. log cot log cos diff. 2 2 2 2 2 2 2 2 2 3 2 2 2 2 3 2 2 2 3 2 2 3 2 2 3 2 3 2 2 3 2 3 2 3 2 3 3 2 3 2 3 3 2 3 3 2 3 3 2 3 diff. Prop. Farts 15.0° 15.1° 15.2° 15.3° 15.4° 15.5° 15.6° 15.7° 15.8° 15.9° 16.0° 16.1° 16.2° 16.3° 16.4° 16.5° 16.6° 16.7° 16.8° 16.9° 17.0° 17.1° 17.2° 17.3° 17.4° 17.5° 17.6° 17.7° 17.8° 17.9° 18.0° 18.1° 18.2° 18.3° 18.4° 18.5° 18.6° 18.7° 18.8° 18,9° 19.0° 19.1° 19.2° 19.3° 19.4° 19.5° 19.6° 19.7° 19.8° 19.9° 20.0° 9.4130 38 38 28 38 27 27 27 27 27 26 27 26 26 26 25 26 25 25 25 25 25 25 34 24 24 24 24 24 23 34 23 23 23 33 23 22 23 32 32 32 22 22 22 21 22 21 23 21 21 21 9.4281 30 30 30 29 30 29 29 29 29 29 28 29 28 28 28 28 27 28 27 27 27 27 27 27 26 27 26 26 26 26 25 26 26 25 25 25 25 25 25 25 24 25 34 24 24 25 23 34 24 34 0.5719 9.9849 75.0° 74.9° 74.8° 74.7° 74.6° 74.5° 74.4° 74.3° 74.2° 74.1° 74.0° 73.9° 73.8° 73.7° 73.6° 73.5° 73.4° 73.3° 73.2° 73.1° 73.0° 72.9° 72.8° 72.7° 72.6° 72.5° 72.4° 72.3° 72.2° 72.1° 72.0° 71.9° 71.8° 71.7° 71.6° 71.5° 71.4° 71.3° 71.2° 71.1° 71.0° 70.9° 70.8° 70.7° 70.6° 70.5° 70.4° 70.3° 70.2° 70.1° 70.0° a bD 3 09 a M w Difference 9.4158 9.4186 9.4214 9.4242 9.4269 9.4296 9.4323 9.4350 9.4377 9.4311 9.4341 9.4371 9.4400 9.4430 9.4459 9.4488 9.4517 9.4546 0.5689 0.5659 0.5629 0.5600 0.5570 0.5541 0.5512 0.5483 0.5454 9.9847 9.9845 9.9843 9.9841 9.9839 9.9837 9.9835 9.9833 9.9831 1 2 3 4 5 6 7 8 9 30 3.0 6.0 9.0 12.0 15.0 18.0 21.0 24.0 27.0 29 2.9 5.8 8.7 11.6 14.5 17.4 20.3 23.2 26.1 9.4403 9.4575 0.5425 9.9828 9.4430 9.4456 9.4482 9.4508 9.4533 9.4559 9.4584 9.4609 9.4634 9.4603 9.4632 9.4660 9.4688 9.4716 9.4744 9.4771 9.4799 9.4826 0.5397 0.5368 0.5340 0.5312 0.5284 0.5256 0.5229 0.5201 0.5174 9.9826 9.9824 9.9822 9.9820 9.9817 9.9815 9.9813 9.9811 9.9808 1 2 3 4 5 6 7 8 9 28 2.8 5.6 8.4 11.2 14.0 16.8 19.6 22.4 25.2 27 2.7 6.4 8.1 10.8 13.5 16.2 18.9 21.6 24.3 9.4659 9.4853 0.5147 9.9806 9.4684 9.4709 9.4733 9.4757 9.4781 9.4805 9.4829 9.4853 9.4876 9.4880 9.4907 9.4934 9.4961 9.4987 9.5014 9.5040 9.5066 9.5092 0.5120 0.5093 0.5066 0.5039 0.5013 0.4986 0.4960 0.4934 0.4908 9.9804 9.9801 9.9799 9.9797 9.9794 9.9792 9.9789 9.9787 9.9785 1 2 3 4 5 6 7 8 9 26 2.6 5.2 7.8 10.4 13.0 15.6 18.2 20.8 23.4 25 2.5 5.0 7.6 10.0 12.5 15.0 17.5 20.0 22.5 9.4900 9.5118 0.4882 9.9782 9.4923 9.4946 9.4969 9.4992 9.5015 9.5037 9.5060 9.5082 9.5104 9.5143 9.5169' 9.5195 9.5220 9.5245 9.5270 9.5295 9.5320 9.5345 0.4857 0.4831 0.4805 0.4780 0.4755 0.4730 0.4705 0.4680 0.4655 9.9780 9.9777 9.9775 9.9772 9.9770 9.9767 9.9764 9.9762 9.9759 1 2 3 4 5 6 7 8 9 24 2.4 4.8 7.2 9.6 12.0 14.4 16.8 19.2 21.6 23 2.3 4.6 6.9 9.2 11.5 13.8 16.1 18.4 20.7 9.5126 9.5370 0.4630 9.9757 9.5148 9.5170 9.5192 9.5213 9.5235 9.5256 9.5278 9.5299 9.5320 9.5394 9.5419 9.5443 9.5467 9.5491 9.5516 9.5539 9.5563 9.5587 0.4606 0.4581 0.4557 0.4533 0.4509 0.4484 0.4461 0.4437 0.4413 9.9754 9.9751 9.9749 9.9746 9.9743 9.9741 9.9738 9.9735 9.9733 1 2 3 4 5 6 7 8 9 22 2.2 4.4 6.6 8.8 11.0 13.2 15.4 17.6 19.8 21 2.1 4.2 6.3 8.4 10.5 12.6 14.7 16.8 18.9 9.5341 9.5611 0.4389 9.9730 log cos difi. log cot com. diff. log tan log sin Angle 70°-75° COSINES, TANGENTS, AND COTANGENTS 33 20°-25° Angle 20.0° 20.1° 20.2° 20.3° 20.4° 20.5° 20.6° 20.7° 20.8° 20.9° 21.0° 21.1° 21.2° 21.3° 21.4° 21.5° 21.6° 21.7° 21.8° 21.9° 22.0° 22.1° 22.2° 22.3° 22.4° 22.5° 22.6° 22.7° 22.8° 22.9° 23.0° 23.1° 23.2° 23.3° 23.4° 23.5° 23.6° 23.7° 23.8° 23.9° 24.0° 24.1° 24.2° 24.3° 24.4° 24.5° 24.6° 24.7° 24.8° 24.9° 25.0° log sin 9.5341 9.5361 9.5382 9.5402 9.5423 9.5443 9.5463 9.5484 9.5504 9.5523 9.5543 9.5563 9.5583 9.5602 9.5621 9.5641 9.5660 9.5679 9.5698 9.5717 9.5736 9.5754 9.5773 9.5792 9.5810 9.5828 9.5847 9.5865 9.5883 9.5901 9.5919 9.5937 9.5954 9.5972 9.5990 9.6007 9.6024 9.6042 9.6059 9.6076 9.6093 9.6110 9.6127 9.6144 9.6161 9.6177 9.6194 9.6210 9.6227 9.6243 9.6259 diS. log tan 9.5611 9.5634 9.5658 9.5681 9.5704 9.5727 9.5750 9.5773 9.5796 9.5819 9.5842 9.5864 9.5887 9.5909 9.5932 9.5954 9.5976 9.5998 9.6020 9.6042 9.6064 9.6086 9.6108 9.6129 9.6151 9.6172 9.6194 9.6215 9.6236 9.6257 9.6279 9.6300 9.6321 9.6341 9.6362 9.6383 9.6404 9.6424 9.6445 9.6465 9.6486 9.6506 9.6527 9.6547 9.6567 9.6587 9.6607 9.6627 9.6647 9.6667 com. difi 9.6687 log cos difi. log cot °.°. m * log tan 22 22 21 22 2] 22 21 21 21 22 21 21 20 21 21 21 20 21 20 21 20 21 20 20 20 20 20 20 log cot 0.4389 0.4366 0.4342 0.4319 0.4296 0.4273 0.4250 0.4227 0.4204 0.4181 0.4158 0.4136 0.4113 0.4091 0.4068 0.4046 0.4024 0.4002 0.3980 0.3958 0.3936 0.3914 0.3892 0.3871 0.3849 0.3828 0.3806 0.3785 0.3764 0.3743 log cos 9.9730 9.9727 9.9724 9.9722 9.9719 9.9716 9.9713 9.9710 9.9707 9.9704 9.9702 9.9699 9.9696 9.9693 9.9690 9.9687 9.9684 9.9681 9.9678 9.9675 9.9672 0.3721 0.3700 0.3679 0.3659 0.3638 0.3617 0.3596 0.3576 0.3555 0.3535 0.3514 0.3494 0.3473 0.3453 0.3433 0.3413 0.3393 0.3373 0.3353 0.3333 0.3313 9.9669 9.9666 9.9662 9.9659 9.9656 9.9653 9.9650 9.9647 9.9643 9.9640 9.9637 9.9634 9.9631 9.9627 9.9624 9.9621 9.9617 9.9614 9.9611 9.9607 9.9604 9.9601 9.9597 9.9594 9.9590 9.9587 9.9583 9.9580 9.9576 9.9573 difi, log sin difi. Angle 70.0° 69.9° 69.8° 69.7° 69.6° 69.5° 69.4° 69.3° 69.2° 69.1° 69.0° 68.9° 68.8° 68.7° 68.6° 68.5° 68.4° 68.3° 68.2° 68.1° 68.0° 67.9° 67.8° 67.7° 67.6° 67.5° 67.4° 67.3° 67.2° 67.1° 67.0° 66.9° 66-8° 66.7° 66.6° 66.5° 66.4° 66.3° 66.2° 66.1° 66.0° 65.9° 65.8° 65.7° 65.6° 65.5° 65.4° 65.3° 65.2° 65.1° 65.0° Prop. Parts Difference 23 2.3 4.6 6.9 9.2 11.6 13.8 16.1 18.4 20.7 21 2.1 4.2 6.3 8.4 10.6 12.6 14.7 16.8 18.9 19 1.9 3.8 5.7 7.6 9.5 11.4 13.3 15.2 17.1 17 1.7 3.4 5.1 6.8 8.5 10.2 11.9 13.6 15.3 2 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 3 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 22 2.2 4.4 6.6 8.8 11.0 13.2 15.4 17.6 19.8 20 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 18 1.8 3.6 5.4 7.2 9.0 10.8 12.6 14.4 16.2 16 1.6 3.2 4.8 6.4 8.0 9.6 11.2 12.8 14.4 4 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 65°-70° 34 TABLE III. LOGARITHMIC SINES 25°-30° Angle log sin difi. log tan com. difi. log cot log cos difi. Prop. Paits 25.0° 25.1° 25.2° 25.3° 9.6259 17 16 16 16 9.6687 19 20 20 19 0.3313 9.9573 4 3 4 4 65.0° 64.9° 64.8° 64.7° +3 'So ■a d Difference 9.6276 9.6292 9.6308 9.6706 9.6726 9.6746 0.3294 0.3274 0.3254 9.9569 9.9566 9.9562 25.4° 25.5° 25.6° 9.6324 9.6340 9.6356 16 16 15 9.6765 9.6785 9.6804 20 19 20 0.3235 0.3215 0.3196 9.9558 9.9555 9.9551 3 4 3 64.6° 64.5° 64.4° w 20 19 25.7° 25.8° 25.9° 26.0° 26.1° 26.2° 26.3° 9.6371 9.6387 9.6403 16 16 15 16 15 16 15 9.6824 9.6843 9.6863 19 20 19 19 19 19 19 0.3176 0.3157 0.3137 9.9548 9.9544 9.9540 4 4 3 4 4 4 3 64.3° 64.2° 64.1° 64.0° 63.9° 63.8° 63.7° 1 2 3 4 5 6 7 8 9 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 1.9 3.8 5.7 7.6 9.5 11.4 13.3 16.2 17.1 9.6418 9.6882 0.3118 9.9537 9.6434 9.6449 9.6465 9.6901 9.6920 9.6939 0.3099 0.3080 0.3061 9.9533 9.9529 9.9525 26.4° 26.5° 26.6° 9.6480 9.6495 9.6510 15 15 16 9.6958 9.6977 9.6996 19 19 19 0.3042 0.3023 0.3004 9.9522 9.9518 9.9514 4 4 4 63.6° 63.5° 63.4° 1 18 1.8 17 1.7 26.7° 26.8° 26.9° 27.0° 27.1° 27.2° 27.3° 9.6526 9.6541 9.6556 9.6570 9.6585 9.6600 9.6615 15 15 14 15 15 15 14 9.7015 9.7034 9.7053 19 19 19 18 19 19 18 0.2985 0.2966 0.2947 9.9510 9.9506 9.9503 4 3 4 4 4 4 4 63.3° 63.2° 63.1° 63.0° 62.9° 62.8° 62.7° 2 3 4 5 6 7 8 9 3.6 5.4 7.2 9.0 10.8 12.6 14.4 16.2 3.4 5.1 6.8 8.5 10.2 11.9 13.6 15.3 9.7072 0.2928 9.9499 9.7090 9.7109 9.7128 0.2910 0.2891 0.2872 9.9495 9.9491 9.9487 27.4° 27.5° 27.6° 9.6629 9.6644 9.6659 15 15 14 9.7146 9.7165 9.7183 19 18 19 0.2854 0.2835 0.2817 9.9483 9.9479 9.9475 4 4 4 62.6° 62.5° 62.4° 1 2 16 1.6 3.2 15 1.5 3.0 27.7° 27.8° 27.9° 28.0° 28.1° 28.2° 28.3° 9.6673 9.6687 9.6702 14 15 14 14 14 15 14 9.7202 9.7220 9.7238 18 18 19 18 18 18 19 0.2798 0.2780 0.2762 9.9471 9.9467 9.9463 4 4 4 4 4 4 4 62.3° 62.2° 62.1° 62.0° 61.9° 61.8° 61.7° 3 4 5 6 7 8 9 4.8 6.4 8.0 9.6 11.2 12.8 14.4 4.5 6.0 7.5 9.0 10.5 12.0 13.5 9.6716 9.7257 0.2743 9.9459 9.6730 9.6744 9.6759 9.7275 9.7293 9.7311 0.2725 0.2707 0.2689 9.9455 9.9451 9.9447 28.4° 28.5° 28.6° 9.6773 9.6787 9.6801 14 14 13 9.7330 9.7348 9.7366 18 18 18 0.2670 0.2652 0.2634 9.9443 9.9439 9.9435 4 4 4 61.6° 61.5° 61.4° 1 2 3 14 1.4 2.8 4.2 13 1.3 2.6 3.9 28.7° 28.8° 28.9° 29.0° 29.1° 29.2° 29.3° 9.6814 9.6828 9.6842 14 14 14 13 14 13 14 9.7384 9.7402 9.7420 18 18 18 17 18 18 18 0.2616 0.2598 0.2580 9.9431 9.9427 9.9422 4 5 4 4 4 4 5 61.3° 61.2° 61.1° 61.0° 60.9° 60.8° 60.7° 4 5 6 7 8 9 5.6 7.0 8.4 9.8 11.2 12.6 5.2 6.5 7.8 9.1 10.4 11.7 9.6856 9.7438 0.2562 9.9418 9.6869 9.6883 9.6896 9.7455 9.7473 9.7491 0.2545 0.2527 0.2509 9.9414 9.9410 9.9406 3 4 29.4° 29.5° 29.6° 29.7° 29.8° 29.9° 30.0° 9.6910 9.6923 9.6937 9.6950 9.6963 9.6977 13 14 13 13 14 13 9.7509 9.7526 9.7544 9.7562 9.7579 9.7597 17 18 18 17 18 17 com. difi. 0.2491 0.2474 0.2456 0.2438 0.2421 0.2403 9.9401 9.9397 9.9393 9.9388 9.9384 9.9380 4 4 5 4 4 5 60.6° 60.5° 60.4° 60.3° 60.2° 60.1° 60.0° 1 2 3 4 5 6 7 8 9 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 9.6990 9.7614 0.2386 9.9375 log cos difi. log cot log tan log sin difi. Angle 60°-65° COSINES, TANGENTS, AND COTANGENTS 35 30°- 35° Angle log sin difi. log tan com. diS. 18 17 18 17 18 17 17 18 17 17 17 17 17 17 17 17 17 17 17 17 17 16 17 17 16 17 16 17 16 17 16 17 16 17 16 17 16 16 17 16 16 16 16 16 16 16 log cot log cos difi. Prop. Parts 30.0° 30.1° 30.2° 30.3° 30.4° 30.5° 30.6° 30.7° 30.8° 30.9° 31.0° 31.1° 31.2° 31.3° 31.4° 31.5° 31.6° 31.7° 31.S° 31.9° 32.0° 32.1° 32.2° 32.3° 32.4° 32.5° 32.6° 32.7° 32.S° 32.9° 33.0° 33.1° 33.2° 33.3° 33.4° 33.5° 33.6° 33.7° 33.8° 33.9-" 34.0° 34.1° 34.2° 34-3° 34.4° 34.5° 34.6° 34.7° 34.8° 34.9° 35.0° 9.6990 13 13 13 13 13 13 12 13 13 12 13 13 12 12 13 12 12 13 12 12 12 12 12 12 12 12 12 12 12 12 11 12 12 12 11 12 11 11 11 11 11 11 11 11 9.7614 0.2386 9.9375 4 4 5 4 5 4 5 4 5 4 5 4 5 5 4 5 5 4 5 5 5 4 5 5 5 5 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 5 5 5 5 6 5 5 5 60.0° 59.9° 59.8° 59.7° 59.6° 59.5° 59.4° 59.3° 59.2° 59.1° 59.0° 58.9° 58.8° 58.7° 58.6° 5S.5° 58.4° 58.3° 58.2° 58.1° 58.0° 57.9° 57.8° 57.7° 57.6° 57.5° 57.4° 57.3° 57.2° 57.1° 57.0° 56.9° 56.8° 56.7° 56.6° 56.5° 56.4° 56.3° 56.2° 56.1° 56.0° 55.9° 55.8° 55.7° 55.6° 55.5° 55.4° 55.3° 55.2° 55.1° 55.0° I ■& ■a 03 w 1 2 3 4 5 6 7 8 9 Difference 9.7003 9.7016 9.7029 9.7042 9.7055 9.7068 9.7080 9.7093 9.7106 9.7632 9.7649 9.7667 9.7684 9.7701 9.7719 9.7736 9.7753 9.7771 0.2368 0.2351 0.2333 0.2316 0.2299 0.2281 0.2264 0.2247 0.2229 9.9371 9.9367 9.9362 9.9358 9.9353 9.9349 9.9344 9.9340 9.9335 18 1.8 3.6 5.4 7.2 9.0 10.8 12.6 14.4 16.2 17 1.7 3.4 5.1 6.8 8.5 10.2 11.9 13.6 15.3 9.7118 9.77S8 0.2212 9.9331 9.7131 9.7144 9.7156 9.7168 9.7181 9.7193 9.7205 9.7218 9.7230 9.7805 9.7822 9.7839 9.7856 9.7S73 9.7S90 9.7907 9.7924 9.7941 0.2195 0.217S 0.2161 0.2144 0.2127 0.2110 0.2093 0.2076 0.2059 9.9326 9.9322 9.9317 9.9312 9.9308 9.9303 9.9298 9.9294 9.9289 1 2 3 4 5 6 7 8 9 16 1.6 3.2 4.8 6.4 S.O 9.6 11.2 12.8 14.4 9.7242 9.795S 0.2042 9.9284 9.7254 9.7266 9.7278 9.7290 9.7302 9.7314 9.7326 9.7338 9-7349 9.7975 9.7992 9.S00S 9.8025 9.S042 9.S0S9 9.8075 9.8092 9.8109 0.2025 0.200S 0.1992 0.1975 0.1958 0.1941 0.1925 0.1908 0.1891 9.9279 9.9275 9.9270 9.9265 9.9260 9.9255 9.9251 9.9246 9.9241 1 2 3 4 5 6 7 8 9 13 1.3 2.6 3.9 5.2 6.5 7.8 9.1 10.4 11.7 12 1.2 2.4 3.6 4.8 6.0 7.2 8.4 9.6 10.8 9.7361 9.S125 0.1875 9.9236 9.7373 9.7384 9.7396 9.7407 9.7419 9.7430 9.7442 9.7453 9.7464 9.8142 9.8158 9.8175 9.8191 9.820S 9.8224 9.8241 9.8257 9.8274 0.1S58 0.1S42 0.1825 0.1809 0.1792 0.1776 0.1759 0.1743 0.1726 9.9231 9.9226 9.9221 9.9216 9.9211 9.9206 9.9201 9.9196 9.9191 1 2 3 4 5 6 7 8 9 11 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 9.7476 9.8290 0.1710 9.9186 9.7487 9.7498 9.7509 9.7520 9.7531 9.7542 9.7553 9.7564 9.7575 9.8306 9.S323 9.S339 9.8355 9.8371 9.83SS 9.8404 9.8420 9.8436 0.1694 0.1677 0.1661 0.1645 0.1629 0.1612 0.1596 0.1 5S0 0.1564 9.9181 9.9175 9.9170 9.9165 9.9160 9.9155 9.9149 9.9144 9.9139 1 2 3 4 5 6 7 8 9 5 0.5 1.0 1.6 2.0 2.5 3.0 3.5 4.0 4.5 6 0.6 1.2 1.8 2.4 3.0 3.6 4.2 4.8 5.4 9.7586 9.8452 0.1548 9.9134 log cos difi. log cot com. difi. log tan log sin difi. Angle 55°-60° 36 TABLE III. LOGARITHMIC SINES 35°-40° Angle log sin difi. log tan com. difi. log cot log COB difi. Prop. Farts 35.0° 35.1° 35.2° 35.3° 9.7586 11 10 n 11 9.8452 16 16 17 16 0.1548 9.9134 6 5 5 6 55.0° 54.9° 54.8° 54.7° 4J •iH bn Difference 9.7597 9.7607 9.7618 9.8468 9.8484 9.8501 0.1532 0.1516 0.1499 9.9128 9.9123 9.9118 35.4° 35.5° 35.6° 9.7629 9.7640 9.7650 n 10 11 9.8517 9.8533 9.8549 16 16 16 0.1483 0.1467 0.1451 9.9112 9.9107 9.9101 5 6 5 54.6° 54.5° 54.4° w 17 16 35.7° 35.8° 35.9° 36.0° 36.1° 36.2° 36.3° 9.7661 9.7671 9.7682 10 11 10 11 10 10 11 9.8565 9.8581 9.8597 16 16 16 16 15 16 16 0.1435 0.1419 0.1403 9.9096 9.9091 9.9085 5 6 5 6 5 6 6 54.3° 54.2° 54.1° 54.0° 53.9° 53.8° 53.7° 1 2 3 4 5 6 7 8 9 1.7 3.4 6.1 6.8 8.5 10.2 11.9 13.6 15.3 1.6 3.2 4.8 6.4 8.0 9.6 11.2 12.8 14.4 9.7692 9.8613 0.1387 9.9080 9.7703 9.7713 9.7723 9.8629 9.8644 9.8660 0.1371 0.1356 0.1340 9.9074 9.9069 9.9063 36.4° 36.5° 36.6° 9.7734 9.7744 9.7754 10 10 10 9.8676 9.8692 9.8708 16 16 16 0.1324 0.1308 0.1292 9.9057 9.9052 9.9046 5 6 5 53.6° 53.5° 53.4° 1 15 1.5 36.7° 36.8° 36.9° 37.0° 37.1° 37.2° 37.3° 9.7764 9.7774 9.7785 10 11 10 10 10 10 10 9.8724 9.8740 9.8755 16 IS 16 16 16 15 16 0.1276 0.1260 0.1245 9.9041 9.9035 9.9029 6 6 6 5 6 6 6 53.3° 53.2° 53.1° 53.0° 52.9° 52.8° 52.7° 2 3 4 5 6 7 8 9 3.0 4.5 6.0 7.5 9.0 10.5 12.0 13.5 9.7795 9.8771 0.1229 9.9023 9.7805 9.7815 9.7825 9.8787 9.8803 9.8818 0.1213 0.1197 0.1182 9.9018 9.9012 9.9006 37.4° 37.5° 37.6° 9.7835 9.7844 9.7854 9 10 10 9.8834 9.8850 9.8865 16 15 16 0.1166 0.1150 0.1135 9.9000 9.8995 9.8989 5 6 6 52.6° 52.5° 52.4° 1 2 11 1.1 2.2 10 1.0 2.0 37.7° 37.8° 37.9° 38.0° 38.1° 38.2° 38.3° 9.7864 9.7874 9.7884 10 10 9 10 10 9 10 9.8881 9.8897 9.8912 16 15 16 16 15 16 15 0.1119 0.1103 0.1088 9.8983 9.8977 9.8971 6 6 6 6 6 6 6 52.3° 52.2° 52.1° 52.0° 51.9° 51.8° 51.7° 3 4 5 6 7 8 9 3.3 4.4 5.5 6.0 7.7 8.8 9.9 3.0 4.0 5.0 6.0 7.0 8.0 9.0 9.7893 9.8928 0.1072 9.8965 9.7903 9.7913 9.7922 9.8944 9.8959 9.8975 0.1056 0.1041 0.1025 9.8959 9.8953 9.8947 38.4° 38.5° 38.6° 9.7932 9.7941 9.7951 9 10 9 9.8990 9.9006 9.9022 16 16 15 0.1010 0.0994 0.0978 9.8941 9.8935 9.8929 6 6 6 51.6° 51.5° 51.4° 1 2 3 9 0.9 1.8 2.7 38.7° 38.8° 38.9° 39.0° 39.1° 39.2° 39.3° 9.7960 9.7970 9.7979 10 9 10 9 9 10 9 9.9037 9.9053 9.9068 16 15 16 15 16 15 16 0.0963 0.0947 0.0932 9.8923 9.8917 9.8911 6 6 6 6 6 6 7 51.3° 51.2° 51.1° 51.0° 50.9° 50.8° 50.7° 4 5 6 7 8 9 3.6 4.5 5.4 6.3 7.2 8.1 9.7989 9.9084 0.0916 9.8905 9.7998 9.8007 9.8017 9.9099 9.9115 9.9130 0.0901 0.0885 0.0870 9.8899 9.8893 9.8887 5 0.5 1.0 1.5 2.0 6 0.6 1.2 1.8 2.4 39.4° 39.5° 39.6° 9.8026 9.8035 9.8044 9 9 9 9.9146 9.9161 9.9176 15 15 16 0.0854 0.0839 0.0824 9.8880 9.S874 9.8868 6 6 6 50.6° 50.5° 50.4° 1 2 3 4 39.7° 39.8° 39.9° 40.0° 9.8053 9.8063 9.8072 10 9 9 9.9192 9.9207 9.9223 15 16 15 0.0808 0.0793 0.0777 9.8862 9.8855 9.8849 7 6 6 50.3° 50.2° 50.1° 50.0° 5 G 7 8 9 2.5 3.0 3.5 4.0 4.5 3.0 3.6 4.2 4.8 5.4 9.8081 9.9238 0.0762 9.8843 log cos dig. log cot com. difi. log tan log sin difi. Angle 50°-55° • COSINES, TANGENTS, AND COTANGENTS 37 40°-45° Angle log sin difi. log tan com. difi. log cot log cos difi. Ifrop. Parts 40.0° 40.1° 40.2° 40.3° 40.4° 40.5° 40.6° 40.7° 40.8° 40.9° 41.0° 41.1° 41.2° 41.3° 41.4° 41.5° 41.6° 41.7° 41.8° 41.9° 42.0° 42.1° 42.2° 42.3° 42.4° 42.5° 42.6° 42.7° 42.8° 42.9° 43.0° 43.1° 43.2° 43.3° 43.4° 43.5° 43.6° 43.7° 43.8° 43.9° 44.0° 44.1° 44.2° 44.3° 44.4° 44.5° 44.6° 44.7° 44.8° 44.9° 45.0° 9.80S1 9.8090 9.8099 9.8108 9.8117 9.8125 9.8134 9.8143 9.8152 9.8161 9.8169 9.8178 9.8187 9.8195 9.S204 9.8213 9.8221 9.8230 9.8238 9.8247 "9^8255 9.8264 9.8272 9.8280 9.8289 9.8297 9.8305 9.8313 9.8322 9.8330 9.8338 9.8346 9.8354 9.8362 9.8370 9.8378 9.8386 9.8394 9.8402 9.8410 9.8418 9.8426 9.8433 9.8441 9.8449 9.S457 9.8464 9.8472 9.8480 9.8487 9.8495 9.9238 9.9254 9.9269 9.9284 9.9300 9.9315 9.9330 9.9346 9.9361 9.9376 9.9392 9.9407 9.9422 9.9438 9.9453 9.9468 9.9483 9.9499 9.9514 9.9529 9.9544 9.9560 9.9575 9.9590 9.9605 9.9621 9.9636 9.9651 9.9666 9.9681 9.9697 9.9712 9.9727 9.9742 9.9757 9.9772 9.9788 9.9803 9.9818 9.9833 9.9848 9.9864 9.9879 9.9894 9.9909 9.9924 9.9939 9.9955 9.9970 9.9985 10.0000 16 15 15 16 15 15 16 15 15 16 15 15 16 15 15 15 16 15 15 15 16 15 15 15 16 15 15 15 15 16 15 15 15 15 15 16 15 15 15 15 16 15 15 15 15 15 16 15 15 15 0.0762 9.8843 0.0746 0.0731 0.0716 0.0700 0.06S5 0.0670 0.0654 0.0639 0.0624 9.8836 9.8830 9.8823 9.8817 9.8810 9.8804 9.8797 9.8791 9.8784 0.0608 9.877S 0.0593 0.0578 0.0562 0.0547 0.0532 0.0517 0.0501 0.04S6 0.0471 9.8771 9.8765 9.8758 9.8751 9.8745 9.8738 9.8731 9.8724 9.8718 0.0456 9.8711 0.0440 0.0425 0.0410 0.0395 0.0379 0.0364 0.0349 0.0334 0.0319 0.0303 0.0288 0.0273 0.0258 0.0243 0.0228 0.0212 0.0197 0.01S2 0.0167 0.0152 0.0136 0.0121 0.0106 0.0091 0.0076 0.0061 0.0045 0.0030 0.0015 0.0000 9.8704 9.8697 9.8690 9.8683 9.8676 9.8669 9.8662 9.8655 9.8648 9.8641 9.8634 9.8627 9.8620 9.8613 9.8606 9.8598 9.8591 9.8584 9.8577 9.8569 9.8562 9.8555 9.8547 9.8540 9.8532 9.8525 9.8517 9.8510 9.8502 9.8495 50.0° 49.9° 49.8° 49.7° 49.6° 49.5° 49.4° 49.3° 49.2° 49.1° 49.0° 48.9° 48.8° 48.7° 48.6° 48.5° 48.4° 48.3° 48.2° 48.1° 48.0° 47.9° 47.8° 47.7° 47.6° 47.5° 47.4° 47.3° 47.2° 47.1° 47.0° 46.9° 46.8° 46.7° 46.6° 46.5° 46.4° 46.3° 46.2° 46.1° 46.0° 45.9° 45. S° 45.7° 45.6° 45.5° 45.4° 45.3° 45.2° 45.1° 45.0° log cos difi. log cot ^' log tan log sin difi. Angle 45°-50° H n 16 1.6 3.2 4.8 6.4 8.0 9.6 11.2 12.8 14.4 15 1.5 3.0 4.5 6.0 7.5 9.0 10.5 12.0 13.5 9 0.9 1.8 2.7 3.6 4.5 5.4 6.3 7.2 8.1 8 0.8 1.6 2.4 33. 4.0 4.8 5.6 6.4 7.2 7 0.7 1.4 2.1 2.8 3.5 4.2 4.9 5.6 6.3 6 0.6 1.2 1.8 2.4 3.0 3.6 4.2 4.8 5.4 38 TABLE OF TRIGONOMETRIC FUNCTIONS TABLE OF NATURAL VALUES OF THE TRIGO- NOMETRIC FUNCTIONS Angle sin COS tan cot sec CSC 0° .0000 1.0000 .0000 00 1.0000 oo 90° 1° .0175 .9998 .0175 57.290 1.0002 57.299 89° 2° .0349 .9994 .0349 28.636 1.0006 28.654 88° 3° .0523 .9986 .0524 19.081 1.0014 19.107 87° 4° .0698 .9976 .0699 14.300 1.0024 14.336 86° 5° .0872 .9962 .0875 11.430 1.0038 11.474 85° 6° .1045 .9945 .1051 9.5144 1.0055 9.5668 84° 7° .1219 .9925 .1228 8.1443 1.0075 8.2055 83° 8° .1392 .9903 .1405 7.1154 1.0098 7.1853 82° 9° .1564 .9877 .1584 6.3138 1.0125 6.3925 81° 10° .1736 .9848 .1763 5.6713 1.0154 5.7588 80° 11° .1908 .9816 .1944 5.1446 1.0187 5.2408 79° 12° .2079 .9781 .2126 4.7046 1.0223 4.8097 78° 13° .2250 .9744 .2309 4.3315 1.0263 4.4454 77° 14° .2419 .9703 .2493 4.0108 1.0306 4.1336 76° 15° .2588 .9659 .2679 3.7321 1.0353 3.8637 75° 16° .2756 .9613 .2S67 3.4874 1.0403 3.6280 74° 17° .2924 .9563 .3057 3^2709 1.0457 3.4203 73° 18° .3090 .9511 .3249 ^3^022$ 1.0515 3.2361 72° 19° .3256 . .9455 .3443 ^9TS42 1.0576 3.0716 71° 20° .3420 .9397 .3640 2.7475 1.0642 2.9238 70° 21° .3584 .9336 .3839 2.6051 1.0711 2.7904 69° 22° .3746 .9272 .4040 2.4751 1.0785 2.6695 68° 23° .3907 .9205 .4245 2.3559 1.0864 2.5593 67° 24° .4067 .9135 .4452 2.2460 1.0946 2.4586 66° 25? .4226 .9063 .4663 2.1445 1.1034 2.3662 65° 64° 26° .4384 .8988 .4877 2.0503 . 1.1126 2.2812 27° .4540 .8910 .5095 1.9626 1.1223 2.2027 63°^ 28° .4695 .8829 .5317 1.8807 1.1326 2.1301 62° 29° .4848 .8746 .5543 1.8040 1.1434 2.0627 61° 30° .5000 .8660 .5774 1.7321 1.1547 2.0000 60° 31° .5150 .8572 .6009 1.6643 1.1666 1.9416 59° 32° .5299 .8480 .6249 1.6003 1.1792 1.8871 58° 33° .5446 JS-587 .6494— — 1.5399- -1,1924 1.8361 57° 34° .5592 .8290 .6745 1.4826 1.2062 1.7883 56° 35" .5736 .S192 .7002 1.4281 1.2208 .1.7434 . -55° 36° .5878 .8090 J265 1.3764 1.2361 1.7013 54° 37° .6018 .7986 <Cj5_3£ 1.3270 1.2521 1.6616 53° 38° .6157 .7S80 .7813 1.2799 1.2690 1.6243 52° 39° .6293 .7771 .8098 1.2349 1.2868 1.5890 51° 40° .6428 .7660 .8391 1.1918 1.3054 1.5557 50° 41° .6561 .7547 .8693 1.1504 1.3250 1.5243 49° 42° .6691 .7431 .9004 1.1106 1.3456 1.4945 48° 43° .6S20 .7314 .9325 1.0724 1.3673 1.4663 47° 44° .6947 .7193 .9657 1.0355 1.3902 1.4396 46° 45° .7071 .7071 1.0000 1.0000 1.4142 1.4142 45° cos sin cot tan CSC sec Angle Date Due JAN I t> 13" 1 I " "W i i •